Class 12 Mathematics Chapter 7 Question Answer | অনুকল | English Medium | ASSEB

Integrals — Questions and Answers

Welcome to HSLC Guru. This lesson gives complete, step-by-step solutions to the textbook exercises of ASSEB Class 12 Mathematics, Chapter 7 — Integrals (অনুকল). Every question of Exercises 7.1 to 7.4 is reproduced exactly as printed and solved in full, so that English-medium students can follow the method as well as the final answer.

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Summary

Integration is the inverse process of differentiation. If $\frac{d}{dx}F(x) = f(x)$, then $F(x)$ is called an anti derivative (or primitive) of $f$, and the whole family of anti derivatives is written as the indefinite integral $\int f(x)\,dx = F(x) + C$, where $C$ is an arbitrary constant of integration. Two basic properties are $\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx$ and $\int k\,f(x)\,dx = k \int f(x)\,dx$.

The standard integrals include $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\ (n \neq -1)$, $\int \cos x\,dx = \sin x + C$, $\int \sin x\,dx = -\cos x + C$, $\int \sec^2 x\,dx = \tan x + C$, $\int e^x\,dx = e^x + C$, $\int \frac{1}{x}\,dx = \log|x| + C$ and $\int a^x\,dx = \frac{a^x}{\log a} + C$. Exercise 7.1 finds anti derivatives by the method of inspection.

The main techniques covered are: integration by substitution (Exercise 7.2), where the variable $x$ is changed to $t$ through $x = g(t)$; integration using trigonometric identities (Exercise 7.3), where products and powers of sine and cosine are rewritten as sums; and the integrals of some particular functions (Exercise 7.4), which use the six standard forms such as $\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$, $\int \frac{dx}{\sqrt{a^2 – x^2}} = \sin^{-1}\frac{x}{a} + C$ and $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log|x + \sqrt{x^2 + a^2}| + C$, together with the technique of completing the square.

Summary: ASSEB Class 12 Mathematics Chapter 7 Integrals explains integration as the reverse of differentiation and develops the indefinite integral. This English-medium guide solves every question of Exercises 7.1, 7.2, 7.3 and 7.4 using inspection, substitution, trigonometric identities and the standard integral forms with completing the square.

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Textbook Questions and Answers

Exercise 7.1

Find an anti derivative (or integral) of the following functions by the method of inspection (Questions 1 to 5).

1. $\sin 2x$

Answer: We look for a function whose derivative is $\sin 2x$. Since $\frac{d}{dx}(\cos 2x) = -2\sin 2x$, we get $\frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x$. Hence an anti derivative is $-\frac{1}{2}\cos 2x$.

2. $\cos 3x$

Answer: Since $\frac{d}{dx}(\sin 3x) = 3\cos 3x$, we have $\frac{d}{dx}\left(\frac{1}{3}\sin 3x\right) = \cos 3x$. Hence an anti derivative is $\frac{1}{3}\sin 3x$.

3. $e^{2x}$

Answer: Since $\frac{d}{dx}(e^{2x}) = 2e^{2x}$, we have $\frac{d}{dx}\left(\frac{1}{2}e^{2x}\right) = e^{2x}$. Hence an anti derivative is $\frac{1}{2}e^{2x}$.

4. $(ax + b)^2$

Answer: Since $\frac{d}{dx}(ax + b)^3 = 3a(ax + b)^2$, we have $\frac{d}{dx}\left[\frac{(ax + b)^3}{3a}\right] = (ax + b)^2$. Hence an anti derivative is $\frac{(ax + b)^3}{3a}$.

5. $\sin 2x – 4e^{3x}$

Answer: Using Questions 1 and 3, an anti derivative is $-\frac{1}{2}\cos 2x – \frac{4}{3}e^{3x}$.

Find the following integrals in Exercises 6 to 20.

6. $\int (4e^{3x} + 1)\,dx$

Answer: $\int (4e^{3x} + 1)\,dx = 4\cdot\frac{e^{3x}}{3} + x + C = \frac{4}{3}e^{3x} + x + C$.

7. $\int x^2\left(1 – \frac{1}{x^2}\right)dx$

Answer: $\int x^2\left(1 – \frac{1}{x^2}\right)dx = \int (x^2 – 1)\,dx = \frac{x^3}{3} – x + C$.

8. $\int (ax^2 + bx + c)\,dx$

Answer: $\int (ax^2 + bx + c)\,dx = \frac{ax^3}{3} + \frac{bx^2}{2} + cx + C$.

9. $\int (2x^2 + e^x)\,dx$

Answer: $\int (2x^2 + e^x)\,dx = \frac{2x^3}{3} + e^x + C$.

10. $\int \left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^2 dx$

Answer: Expanding, $\left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^2 = x – 2 + \frac{1}{x}$. Therefore

$$\int \left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^2 dx = \frac{x^2}{2} – 2x + \log|x| + C$$

11. $\int \frac{x^3 + 5x^2 – 4}{x^2}\,dx$

Answer: $\frac{x^3 + 5x^2 – 4}{x^2} = x + 5 – 4x^{-2}$. Therefore

$$\int \frac{x^3 + 5x^2 – 4}{x^2}\,dx = \frac{x^2}{2} + 5x + \frac{4}{x} + C$$

12. $\int \frac{x^3 + 3x + 4}{\sqrt{x}}\,dx$

Answer: $\frac{x^3 + 3x + 4}{\sqrt{x}} = x^{5/2} + 3x^{1/2} + 4x^{-1/2}$. Therefore

$$\int \frac{x^3 + 3x + 4}{\sqrt{x}}\,dx = \frac{2}{7}x^{7/2} + 2x^{3/2} + 8\sqrt{x} + C$$

13. $\int \frac{x^3 – x^2 + x – 1}{x – 1}\,dx$

Answer: Factorising the numerator, $x^3 – x^2 + x – 1 = x^2(x – 1) + (x – 1) = (x – 1)(x^2 + 1)$. So the integrand is $x^2 + 1$, and

$$\int \frac{x^3 – x^2 + x – 1}{x – 1}\,dx = \int (x^2 + 1)\,dx = \frac{x^3}{3} + x + C$$

14. $\int (1 – x)\sqrt{x}\,dx$

Answer: $(1 – x)\sqrt{x} = x^{1/2} – x^{3/2}$. Therefore

$$\int (1 – x)\sqrt{x}\,dx = \frac{2}{3}x^{3/2} – \frac{2}{5}x^{5/2} + C$$

15. $\int \sqrt{x}\,(3x^2 + 2x + 3)\,dx$

Answer: $\sqrt{x}\,(3x^2 + 2x + 3) = 3x^{5/2} + 2x^{3/2} + 3x^{1/2}$. Therefore

$$\int \sqrt{x}\,(3x^2 + 2x + 3)\,dx = \frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C$$

16. $\int (2x – 3\cos x + e^x)\,dx$

Answer: $\int (2x – 3\cos x + e^x)\,dx = x^2 – 3\sin x + e^x + C$.

17. $\int (2x^2 – 3\sin x + 5\sqrt{x})\,dx$

Answer: $\int (2x^2 – 3\sin x + 5\sqrt{x})\,dx = \frac{2}{3}x^3 + 3\cos x + \frac{10}{3}x^{3/2} + C$.

18. $\int \sec x\,(\sec x + \tan x)\,dx$

Answer: $\sec x\,(\sec x + \tan x) = \sec^2 x + \sec x \tan x$. Therefore $\int \sec x\,(\sec x + \tan x)\,dx = \tan x + \sec x + C$.

19. $\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx$

Answer: $\frac{\sec^2 x}{\operatorname{cosec}^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x = \sec^2 x – 1$. Therefore

$$\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx = \int (\sec^2 x – 1)\,dx = \tan x – x + C$$

20. $\int \frac{2 – 3\sin x}{\cos^2 x}\,dx$

Answer: $\frac{2 – 3\sin x}{\cos^2 x} = 2\sec^2 x – 3\,\frac{\sin x}{\cos^2 x} = 2\sec^2 x – 3\sec x \tan x$. Therefore

$$\int \frac{2 – 3\sin x}{\cos^2 x}\,dx = 2\tan x – 3\sec x + C$$

Choose the correct answer in Exercises 21 and 22.

21. The anti derivative of $\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)$ equals

(A) $\frac{1}{3}x^{1/3} + 2x^{1/2} + C$    (B) $\frac{2}{3}x^{2/3} + \frac{1}{2}x^2 + C$    (C) $\frac{2}{3}x^{3/2} + 2x^{1/2} + C$    (D) $\frac{3}{2}x^{3/2} + \frac{1}{2}x^{1/2} + C$

Answer: $\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)dx = \int (x^{1/2} + x^{-1/2})\,dx = \frac{2}{3}x^{3/2} + 2x^{1/2} + C$. Hence the correct option is (C).

22. If $\frac{d}{dx}f(x) = 4x^3 – \frac{3}{x^4}$ such that $f(2) = 0$, then $f(x)$ is

(A) $x^4 + \frac{1}{x^3} – \frac{129}{8}$    (B) $x^3 + \frac{1}{x^4} + \frac{129}{8}$    (C) $x^4 + \frac{1}{x^3} + \frac{129}{8}$    (D) $x^3 + \frac{1}{x^4} – \frac{129}{8}$

Answer: Integrating, $f(x) = \int \left(4x^3 – 3x^{-4}\right)dx = x^4 + \frac{1}{x^3} + C$. The condition $f(2) = 0$ gives $16 + \frac{1}{8} + C = 0$, so $C = -\frac{129}{8}$. Hence $f(x) = x^4 + \frac{1}{x^3} – \frac{129}{8}$, and the correct option is (A).

Exercise 7.2

Integrate the functions in Exercises 1 to 37.

1. $\frac{2x}{1 + x^2}$

Answer: Put $t = 1 + x^2$, so $dt = 2x\,dx$. Then $\int \frac{2x}{1 + x^2}\,dx = \int \frac{dt}{t} = \log|t| + C = \log(1 + x^2) + C$.

2. $\frac{(\log x)^2}{x}$

Answer: Put $t = \log x$, so $dt = \frac{1}{x}\,dx$. Then $\int \frac{(\log x)^2}{x}\,dx = \int t^2\,dt = \frac{t^3}{3} + C = \frac{(\log x)^3}{3} + C$.

3. $\frac{1}{x + x\log x}$

Answer: $\frac{1}{x + x\log x} = \frac{1}{x(1 + \log x)}$. Put $t = 1 + \log x$, so $dt = \frac{1}{x}\,dx$. Then $\int \frac{dt}{t} = \log|t| + C = \log|1 + \log x| + C$.

4. $\sin x\,\sin(\cos x)$

Answer: Put $t = \cos x$, so $dt = -\sin x\,dx$. Then $\int \sin(\cos x)\,\sin x\,dx = -\int \sin t\,dt = \cos t + C = \cos(\cos x) + C$.

5. $\sin(ax + b)\cos(ax + b)$

Answer: Put $t = \sin(ax + b)$, so $dt = a\cos(ax + b)\,dx$. Then

$$\int \sin(ax + b)\cos(ax + b)\,dx = \frac{1}{a}\int t\,dt = \frac{t^2}{2a} + C = \frac{\sin^2(ax + b)}{2a} + C$$

6. $\sqrt{ax + b}$

Answer: Put $t = ax + b$, so $dt = a\,dx$. Then $\int \sqrt{ax + b}\,dx = \frac{1}{a}\int t^{1/2}\,dt = \frac{1}{a}\cdot\frac{2}{3}t^{3/2} + C = \frac{2}{3a}(ax + b)^{3/2} + C$.

7. $x\sqrt{x + 2}$

Answer: Put $t = x + 2$, so $x = t – 2$ and $dx = dt$. Then $\int x\sqrt{x + 2}\,dx = \int (t – 2)t^{1/2}\,dt = \int (t^{3/2} – 2t^{1/2})\,dt = \frac{2}{5}t^{5/2} – \frac{4}{3}t^{3/2} + C$.

$$\int x\sqrt{x + 2}\,dx = \frac{2}{5}(x + 2)^{5/2} – \frac{4}{3}(x + 2)^{3/2} + C$$

8. $x\sqrt{1 + 2x^2}$

Answer: Put $t = 1 + 2x^2$, so $dt = 4x\,dx$. Then $\int x\sqrt{1 + 2x^2}\,dx = \frac{1}{4}\int t^{1/2}\,dt = \frac{1}{4}\cdot\frac{2}{3}t^{3/2} + C = \frac{1}{6}(1 + 2x^2)^{3/2} + C$.

9. $(4x + 2)\sqrt{x^2 + x + 1}$

Answer: Note $\frac{d}{dx}(x^2 + x + 1) = 2x + 1$, and $4x + 2 = 2(2x + 1)$. Put $t = x^2 + x + 1$, so $dt = (2x + 1)\,dx$. Then $\int (4x + 2)\sqrt{x^2 + x + 1}\,dx = 2\int t^{1/2}\,dt = \frac{4}{3}t^{3/2} + C = \frac{4}{3}(x^2 + x + 1)^{3/2} + C$.

10. $\frac{1}{x – \sqrt{x}}$

Answer: Put $t = \sqrt{x}$, so $x = t^2$ and $dx = 2t\,dt$. Then $\int \frac{1}{x – \sqrt{x}}\,dx = \int \frac{2t\,dt}{t^2 – t} = \int \frac{2t\,dt}{t(t – 1)} = \int \frac{2\,dt}{t – 1} = 2\log|t – 1| + C = 2\log|\sqrt{x} – 1| + C$.

11. $\frac{x}{\sqrt{x + 4}}$, $x > 0$

Answer: Put $t = x + 4$, so $x = t – 4$ and $dx = dt$. Then $\int \frac{x}{\sqrt{x + 4}}\,dx = \int \frac{t – 4}{\sqrt{t}}\,dt = \int (t^{1/2} – 4t^{-1/2})\,dt = \frac{2}{3}t^{3/2} – 8t^{1/2} + C$.

$$\int \frac{x}{\sqrt{x + 4}}\,dx = \frac{2}{3}(x + 4)^{3/2} – 8\sqrt{x + 4} + C$$

12. $(x^3 – 1)^{1/3}\,x^5$

Answer: Put $t = x^3 – 1$, so $x^3 = t + 1$ and $dt = 3x^2\,dx$. Since $x^5\,dx = x^3\cdot x^2\,dx = (t + 1)\frac{dt}{3}$,

$$\int (x^3 – 1)^{1/3}x^5\,dx = \frac{1}{3}\int t^{1/3}(t + 1)\,dt = \frac{1}{3}\int (t^{4/3} + t^{1/3})\,dt = \frac{1}{7}(x^3 – 1)^{7/3} + \frac{1}{4}(x^3 – 1)^{4/3} + C$$

13. $\frac{x^2}{(2 + 3x^3)^3}$

Answer: Put $t = 2 + 3x^3$, so $dt = 9x^2\,dx$. Then $\int \frac{x^2}{(2 + 3x^3)^3}\,dx = \frac{1}{9}\int t^{-3}\,dt = \frac{1}{9}\cdot\frac{t^{-2}}{-2} + C = -\frac{1}{18(2 + 3x^3)^2} + C$.

14. $\frac{1}{x(\log x)^m}$, $x > 0$, $m \neq 1$

Answer: Put $t = \log x$, so $dt = \frac{1}{x}\,dx$. Then $\int \frac{1}{x(\log x)^m}\,dx = \int t^{-m}\,dt = \frac{t^{1-m}}{1 – m} + C = \frac{(\log x)^{1-m}}{1 – m} + C$.

15. $\frac{x}{9 – 4x^2}$

Answer: Put $t = 9 – 4x^2$, so $dt = -8x\,dx$. Then $\int \frac{x}{9 – 4x^2}\,dx = -\frac{1}{8}\int \frac{dt}{t} = -\frac{1}{8}\log|t| + C = -\frac{1}{8}\log|9 – 4x^2| + C$.

16. $e^{2x + 3}$

Answer: Put $t = 2x + 3$, so $dt = 2\,dx$. Then $\int e^{2x + 3}\,dx = \frac{1}{2}\int e^t\,dt = \frac{1}{2}e^{2x + 3} + C$.

17. $\frac{x}{e^{x^2}}$

Answer: $\frac{x}{e^{x^2}} = x\,e^{-x^2}$. Put $t = -x^2$, so $dt = -2x\,dx$. Then $\int x\,e^{-x^2}\,dx = -\frac{1}{2}\int e^t\,dt = -\frac{1}{2}e^{-x^2} + C$.

18. $\frac{e^{\tan^{-1} x}}{1 + x^2}$

Answer: Put $t = \tan^{-1} x$, so $dt = \frac{1}{1 + x^2}\,dx$. Then $\int \frac{e^{\tan^{-1} x}}{1 + x^2}\,dx = \int e^t\,dt = e^t + C = e^{\tan^{-1} x} + C$.

19. $\frac{e^{2x} – 1}{e^{2x} + 1}$

Answer: Dividing numerator and denominator by $e^x$, $\frac{e^{2x} – 1}{e^{2x} + 1} = \frac{e^x – e^{-x}}{e^x + e^{-x}}$. Put $t = e^x + e^{-x}$, so $dt = (e^x – e^{-x})\,dx$. Then $\int \frac{e^x – e^{-x}}{e^x + e^{-x}}\,dx = \int \frac{dt}{t} = \log|t| + C = \log(e^x + e^{-x}) + C$.

20. $\frac{e^{2x} – e^{-2x}}{e^{2x} + e^{-2x}}$

Answer: Put $t = e^{2x} + e^{-2x}$, so $dt = 2(e^{2x} – e^{-2x})\,dx$. Then $\int \frac{e^{2x} – e^{-2x}}{e^{2x} + e^{-2x}}\,dx = \frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}\log(e^{2x} + e^{-2x}) + C$.

21. $\tan^2(2x – 3)$

Answer: $\tan^2(2x – 3) = \sec^2(2x – 3) – 1$. Therefore

$$\int \tan^2(2x – 3)\,dx = \int [\sec^2(2x – 3) – 1]\,dx = \frac{1}{2}\tan(2x – 3) – x + C$$

22. $\sec^2(7 – 4x)$

Answer: Put $t = 7 – 4x$, so $dt = -4\,dx$. Then $\int \sec^2(7 – 4x)\,dx = -\frac{1}{4}\int \sec^2 t\,dt = -\frac{1}{4}\tan(7 – 4x) + C$.

23. $\frac{\sin^{-1} x}{\sqrt{1 – x^2}}$

Answer: Put $t = \sin^{-1} x$, so $dt = \frac{1}{\sqrt{1 – x^2}}\,dx$. Then $\int \frac{\sin^{-1} x}{\sqrt{1 – x^2}}\,dx = \int t\,dt = \frac{t^2}{2} + C = \frac{(\sin^{-1} x)^2}{2} + C$.

24. $\frac{2\cos x – 3\sin x}{6\cos x + 4\sin x}$

Answer: Observe that $\frac{d}{dx}(6\cos x + 4\sin x) = -6\sin x + 4\cos x = 2(2\cos x – 3\sin x)$. Hence $2\cos x – 3\sin x = \frac{1}{2}\frac{d}{dx}(6\cos x + 4\sin x)$, so

$$\int \frac{2\cos x – 3\sin x}{6\cos x + 4\sin x}\,dx = \frac{1}{2}\log|6\cos x + 4\sin x| + C$$

25. $\frac{1}{\cos^2 x\,(1 – \tan x)^2}$

Answer: $\frac{1}{\cos^2 x\,(1 – \tan x)^2} = \frac{\sec^2 x}{(1 – \tan x)^2}$. Put $t = 1 – \tan x$, so $dt = -\sec^2 x\,dx$. Then $\int \frac{\sec^2 x}{(1 – \tan x)^2}\,dx = -\int t^{-2}\,dt = \frac{1}{t} + C = \frac{1}{1 – \tan x} + C$.

26. $\frac{\cos \sqrt{x}}{\sqrt{x}}$

Answer: Put $t = \sqrt{x}$, so $dt = \frac{1}{2\sqrt{x}}\,dx$, i.e. $\frac{dx}{\sqrt{x}} = 2\,dt$. Then $\int \frac{\cos \sqrt{x}}{\sqrt{x}}\,dx = 2\int \cos t\,dt = 2\sin t + C = 2\sin \sqrt{x} + C$.

27. $\sqrt{\sin 2x}\,\cos 2x$

Answer: Put $t = \sin 2x$, so $dt = 2\cos 2x\,dx$. Then $\int \sqrt{\sin 2x}\,\cos 2x\,dx = \frac{1}{2}\int t^{1/2}\,dt = \frac{1}{2}\cdot\frac{2}{3}t^{3/2} + C = \frac{1}{3}(\sin 2x)^{3/2} + C$.

28. $\frac{\cos x}{\sqrt{1 + \sin x}}$

Answer: Put $t = 1 + \sin x$, so $dt = \cos x\,dx$. Then $\int \frac{\cos x}{\sqrt{1 + \sin x}}\,dx = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} + C = 2\sqrt{1 + \sin x} + C$.

29. $\cot x\,\log \sin x$

Answer: Put $t = \log \sin x$, so $dt = \frac{\cos x}{\sin x}\,dx = \cot x\,dx$. Then $\int \cot x\,\log \sin x\,dx = \int t\,dt = \frac{t^2}{2} + C = \frac{(\log \sin x)^2}{2} + C$.

30. $\frac{\sin x}{1 + \cos x}$

Answer: Put $t = 1 + \cos x$, so $dt = -\sin x\,dx$. Then $\int \frac{\sin x}{1 + \cos x}\,dx = -\int \frac{dt}{t} = -\log|t| + C = -\log|1 + \cos x| + C$.

31. $\frac{\sin x}{(1 + \cos x)^2}$

Answer: Put $t = 1 + \cos x$, so $dt = -\sin x\,dx$. Then $\int \frac{\sin x}{(1 + \cos x)^2}\,dx = -\int t^{-2}\,dt = \frac{1}{t} + C = \frac{1}{1 + \cos x} + C$.

32. $\frac{1}{1 + \cot x}$

Answer: $\frac{1}{1 + \cot x} = \frac{\sin x}{\sin x + \cos x}$. Writing $\sin x = \frac{1}{2}[(\sin x + \cos x) + (\sin x – \cos x)]$ and noting $\frac{d}{dx}(\sin x + \cos x) = \cos x – \sin x$,

$$\int \frac{\sin x}{\sin x + \cos x}\,dx = \frac{1}{2}\int \left[1 – \frac{\cos x – \sin x}{\sin x + \cos x}\right]dx = \frac{x}{2} – \frac{1}{2}\log|\sin x + \cos x| + C$$

33. $\frac{1}{1 – \tan x}$

Answer: $\frac{1}{1 – \tan x} = \frac{\cos x}{\cos x – \sin x}$. Writing $\cos x = \frac{1}{2}[(\cos x – \sin x) + (\cos x + \sin x)]$ and noting $\frac{d}{dx}(\cos x – \sin x) = -(\sin x + \cos x)$,

$$\int \frac{\cos x}{\cos x – \sin x}\,dx = \frac{1}{2}\int \left[1 + \frac{\cos x + \sin x}{\cos x – \sin x}\right]dx = \frac{x}{2} – \frac{1}{2}\log|\cos x – \sin x| + C$$

34. $\frac{\sqrt{\tan x}}{\sin x\,\cos x}$

Answer: Since $\sin x\,\cos x = \cos^2 x\,\tan x$, we have $\frac{\sqrt{\tan x}}{\sin x\,\cos x} = \frac{\sec^2 x}{\sqrt{\tan x}}$. Put $t = \tan x$, so $dt = \sec^2 x\,dx$. Then $\int \frac{\sec^2 x}{\sqrt{\tan x}}\,dx = \int t^{-1/2}\,dt = 2\sqrt{t} + C = 2\sqrt{\tan x} + C$.

35. $\frac{(1 + \log x)^2}{x}$

Answer: Put $t = 1 + \log x$, so $dt = \frac{1}{x}\,dx$. Then $\int \frac{(1 + \log x)^2}{x}\,dx = \int t^2\,dt = \frac{t^3}{3} + C = \frac{(1 + \log x)^3}{3} + C$.

36. $\frac{(x + 1)(x + \log x)^2}{x}$

Answer: Since $\frac{d}{dx}(x + \log x) = 1 + \frac{1}{x} = \frac{x + 1}{x}$, put $t = x + \log x$, so $dt = \frac{x + 1}{x}\,dx$. Then $\int \frac{(x + 1)(x + \log x)^2}{x}\,dx = \int t^2\,dt = \frac{t^3}{3} + C = \frac{(x + \log x)^3}{3} + C$.

37. $\frac{x^3 \sin(\tan^{-1} x^4)}{1 + x^8}$

Answer: Put $t = \tan^{-1}(x^4)$, so $dt = \frac{4x^3}{1 + x^8}\,dx$, i.e. $\frac{x^3}{1 + x^8}\,dx = \frac{dt}{4}$. Then $\int \frac{x^3 \sin(\tan^{-1} x^4)}{1 + x^8}\,dx = \frac{1}{4}\int \sin t\,dt = -\frac{1}{4}\cos t + C = -\frac{1}{4}\cos(\tan^{-1} x^4) + C$.

Choose the correct answer in Exercises 38 and 39.

38. $\int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x}\,dx$ equals

(A) $10^x – x^{10} + C$    (B) $10^x + x^{10} + C$    (C) $(10^x – x^{10})^{-1} + C$    (D) $\log(10^x + x^{10}) + C$

Answer: Note $\frac{d}{dx}(x^{10} + 10^x) = 10x^9 + 10^x \log_e 10$, which is exactly the numerator. Hence the integrand is $\frac{f^{\prime}(x)}{f(x)}$ with $f(x) = x^{10} + 10^x$, and the integral is $\log(x^{10} + 10^x) + C$. The correct option is (D).

39. $\int \frac{dx}{\sin^2 x\,\cos^2 x}$ equals

(A) $\tan x + \cot x + C$    (B) $\tan x – \cot x + C$    (C) $\tan x\,\cot x + C$    (D) $\tan x – \cot 2x + C$

Answer: $\frac{1}{\sin^2 x\,\cos^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x\,\cos^2 x} = \sec^2 x + \operatorname{cosec}^2 x$. Therefore $\int \frac{dx}{\sin^2 x\,\cos^2 x} = \tan x – \cot x + C$. The correct option is (B).

Exercise 7.3

Find the integrals of the functions in Exercises 1 to 22.

1. $\sin^2(2x + 5)$

Answer: $\sin^2(2x + 5) = \frac{1 – \cos(4x + 10)}{2}$. Therefore

$$\int \sin^2(2x + 5)\,dx = \frac{1}{2}\int [1 – \cos(4x + 10)]\,dx = \frac{x}{2} – \frac{1}{8}\sin(4x + 10) + C$$

2. $\sin 3x\,\cos 4x$

Answer: Using $\sin A\cos B = \frac{1}{2}[\sin(A + B) + \sin(A – B)]$, $\sin 3x\cos 4x = \frac{1}{2}[\sin 7x – \sin x]$. Therefore

$$\int \sin 3x\,\cos 4x\,dx = \frac{1}{2}\left[-\frac{\cos 7x}{7} + \cos x\right] + C = -\frac{1}{14}\cos 7x + \frac{1}{2}\cos x + C$$

3. $\cos 2x\,\cos 4x\,\cos 6x$

Answer: $\cos 2x\cos 4x = \frac{1}{2}(\cos 6x + \cos 2x)$, so $\cos 2x\cos 4x\cos 6x = \frac{1}{2}(\cos^2 6x + \cos 2x\cos 6x)$. Using $\cos^2 6x = \frac{1 + \cos 12x}{2}$ and $\cos 2x\cos 6x = \frac{1}{2}(\cos 8x + \cos 4x)$, this equals $\frac{1}{4}(1 + \cos 12x + \cos 8x + \cos 4x)$. Therefore

$$\int \cos 2x\cos 4x\cos 6x\,dx = \frac{1}{4}\left[x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4}\right] + C$$

4. $\sin^3(2x + 1)$

Answer: Using $\sin^3\theta = \frac{3\sin\theta – \sin 3\theta}{4}$ with $\theta = 2x + 1$, $\sin^3(2x + 1) = \frac{1}{4}[3\sin(2x + 1) – \sin(6x + 3)]$. Therefore

$$\int \sin^3(2x + 1)\,dx = -\frac{3}{8}\cos(2x + 1) + \frac{1}{24}\cos(6x + 3) + C$$

5. $\sin^3 x\,\cos^3 x$

Answer: Write $\sin^3 x\cos^3 x = \sin^3 x\,(1 – \sin^2 x)\cos x$. Put $t = \sin x$, so $dt = \cos x\,dx$. Then $\int t^3(1 – t^2)\,dt = \int (t^3 – t^5)\,dt = \frac{t^4}{4} – \frac{t^6}{6} + C$.

$$\int \sin^3 x\,\cos^3 x\,dx = \frac{\sin^4 x}{4} – \frac{\sin^6 x}{6} + C$$

6. $\sin x\,\sin 2x\,\sin 3x$

Answer: $\sin x\sin 3x = \frac{1}{2}(\cos 2x – \cos 4x)$, so $\sin x\sin 2x\sin 3x = \frac{1}{2}\sin 2x(\cos 2x – \cos 4x) = \frac{1}{4}(\sin 4x – \sin 6x + \sin 2x)$. Therefore

$$\int \sin x\sin 2x\sin 3x\,dx = -\frac{1}{16}\cos 4x + \frac{1}{24}\cos 6x – \frac{1}{8}\cos 2x + C$$

7. $\sin 4x\,\sin 8x$

Answer: $\sin 4x\sin 8x = \frac{1}{2}(\cos 4x – \cos 12x)$. Therefore

$$\int \sin 4x\sin 8x\,dx = \frac{1}{2}\left[\frac{\sin 4x}{4} – \frac{\sin 12x}{12}\right] + C = \frac{1}{8}\sin 4x – \frac{1}{24}\sin 12x + C$$

8. $\frac{1 – \cos x}{1 + \cos x}$

Answer: Using $1 – \cos x = 2\sin^2\frac{x}{2}$ and $1 + \cos x = 2\cos^2\frac{x}{2}$, $\frac{1 – \cos x}{1 + \cos x} = \tan^2\frac{x}{2} = \sec^2\frac{x}{2} – 1$. Therefore

$$\int \frac{1 – \cos x}{1 + \cos x}\,dx = 2\tan\frac{x}{2} – x + C$$

9. $\frac{\cos x}{1 + \cos x}$

Answer: $\frac{\cos x}{1 + \cos x} = 1 – \frac{1}{1 + \cos x} = 1 – \frac{1}{2}\sec^2\frac{x}{2}$. Therefore

$$\int \frac{\cos x}{1 + \cos x}\,dx = x – \tan\frac{x}{2} + C$$

10. $\sin^4 x$

Answer: $\sin^4 x = \left(\frac{1 – \cos 2x}{2}\right)^2 = \frac{1}{4}(1 – 2\cos 2x + \cos^2 2x) = \frac{3}{8} – \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$. Therefore

$$\int \sin^4 x\,dx = \frac{3x}{8} – \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$$

11. $\cos^4 2x$

Answer: $\cos^4 2x = \left(\frac{1 + \cos 4x}{2}\right)^2 = \frac{1}{4}(1 + 2\cos 4x + \cos^2 4x) = \frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x$. Therefore

$$\int \cos^4 2x\,dx = \frac{3x}{8} + \frac{1}{8}\sin 4x + \frac{1}{64}\sin 8x + C$$

12. $\frac{\sin^2 x}{1 + \cos x}$

Answer: $\frac{\sin^2 x}{1 + \cos x} = \frac{1 – \cos^2 x}{1 + \cos x} = \frac{(1 – \cos x)(1 + \cos x)}{1 + \cos x} = 1 – \cos x$. Therefore $\int \frac{\sin^2 x}{1 + \cos x}\,dx = x – \sin x + C$.

13. $\frac{\cos 2x – \cos 2\alpha}{\cos x – \cos \alpha}$

Answer: Using $\cos 2x = 2\cos^2 x – 1$ and $\cos 2\alpha = 2\cos^2\alpha – 1$, the numerator becomes $2(\cos^2 x – \cos^2\alpha) = 2(\cos x – \cos\alpha)(\cos x + \cos\alpha)$. Hence the integrand equals $2(\cos x + \cos\alpha)$, and

$$\int \frac{\cos 2x – \cos 2\alpha}{\cos x – \cos \alpha}\,dx = 2(\sin x + x\cos\alpha) + C$$

14. $\frac{\cos x – \sin x}{1 + \sin 2x}$

Answer: Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x\cos x = (\sin x + \cos x)^2$, put $t = \sin x + \cos x$, so $dt = (\cos x – \sin x)\,dx$. Then $\int \frac{\cos x – \sin x}{(\sin x + \cos x)^2}\,dx = \int \frac{dt}{t^2} = -\frac{1}{t} + C = -\frac{1}{\sin x + \cos x} + C$.

15. $\tan^3 2x\,\sec 2x$

Answer: $\tan^3 2x\sec 2x = (\sec^2 2x – 1)\tan 2x\sec 2x$. Put $t = \sec 2x$, so $dt = 2\sec 2x\tan 2x\,dx$. Then $\int (\sec^2 2x – 1)\sec 2x\tan 2x\,dx = \frac{1}{2}\int (t^2 – 1)\,dt = \frac{t^3}{6} – \frac{t}{2} + C$.

$$\int \tan^3 2x\,\sec 2x\,dx = \frac{1}{6}\sec^3 2x – \frac{1}{2}\sec 2x + C$$

16. $\tan^4 x$

Answer: $\tan^4 x = \tan^2 x(\sec^2 x – 1) = \tan^2 x\sec^2 x – (\sec^2 x – 1)$. With $t = \tan x$, $\int \tan^2 x\sec^2 x\,dx = \frac{\tan^3 x}{3}$, and $\int (\sec^2 x – 1)\,dx = \tan x – x$. Therefore

$$\int \tan^4 x\,dx = \frac{\tan^3 x}{3} – \tan x + x + C$$

17. $\frac{\sin^3 x + \cos^3 x}{\sin^2 x\,\cos^2 x}$

Answer: Splitting, $\frac{\sin^3 x + \cos^3 x}{\sin^2 x\cos^2 x} = \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} = \sec x\tan x + \operatorname{cosec} x\cot x$. Therefore

$$\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x\,\cos^2 x}\,dx = \sec x – \operatorname{cosec} x + C$$

18. $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$

Answer: Since $\cos 2x = 1 – 2\sin^2 x$, the numerator $\cos 2x + 2\sin^2 x = 1$. So the integrand is $\frac{1}{\cos^2 x} = \sec^2 x$, and $\int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx = \tan x + C$.

19. $\frac{1}{\sin x\,\cos^3 x}$

Answer: Dividing numerator and denominator by $\cos^4 x$, $\frac{1}{\sin x\cos^3 x} = \frac{\sec^4 x}{\tan x} = \frac{(1 + \tan^2 x)\sec^2 x}{\tan x}$. Put $t = \tan x$, so $dt = \sec^2 x\,dx$. Then $\int \frac{1 + t^2}{t}\,dt = \int \left(\frac{1}{t} + t\right)dt = \log|t| + \frac{t^2}{2} + C$.

$$\int \frac{1}{\sin x\,\cos^3 x}\,dx = \log|\tan x| + \frac{\tan^2 x}{2} + C$$

20. $\frac{\cos 2x}{(\cos x + \sin x)^2}$

Answer: Since $\cos 2x = \cos^2 x – \sin^2 x = (\cos x – \sin x)(\cos x + \sin x)$, the integrand equals $\frac{\cos x – \sin x}{\cos x + \sin x}$. Put $t = \cos x + \sin x$, so $dt = (\cos x – \sin x)\,dx$. Then $\int \frac{dt}{t} = \log|t| + C = \log|\cos x + \sin x| + C$.

21. $\sin^{-1}(\cos x)$

Answer: Since $\cos x = \sin\left(\frac{\pi}{2} – x\right)$, we have $\sin^{-1}(\cos x) = \frac{\pi}{2} – x$. Therefore

$$\int \sin^{-1}(\cos x)\,dx = \int \left(\frac{\pi}{2} – x\right)dx = \frac{\pi x}{2} – \frac{x^2}{2} + C$$

22. $\frac{1}{\cos(x – a)\cos(x – b)}$

Answer: Multiply and divide by $\sin(a – b)$. Since $\sin(a – b) = \sin[(x – b) – (x – a)] = \sin(x – b)\cos(x – a) – \cos(x – b)\sin(x – a)$,

$$\frac{1}{\cos(x – a)\cos(x – b)} = \frac{1}{\sin(a – b)}\,[\tan(x – b) – \tan(x – a)]$$

Integrating and using $\int \tan\theta\,d\theta = -\log|\cos\theta|$,

$$\int \frac{dx}{\cos(x – a)\cos(x – b)} = \frac{1}{\sin(a – b)}\,\log\left|\frac{\cos(x – a)}{\cos(x – b)}\right| + C$$

Choose the correct answer in Exercises 23 and 24.

23. $\int \frac{\sin^2 x – \cos^2 x}{\sin^2 x\,\cos^2 x}\,dx$ is equal to

(A) $\tan x + \cot x + C$    (B) $\tan x + \operatorname{cosec} x + C$    (C) $-\tan x + \cot x + C$    (D) $\tan x + \sec x + C$

Answer: $\frac{\sin^2 x – \cos^2 x}{\sin^2 x\cos^2 x} = \frac{1}{\cos^2 x} – \frac{1}{\sin^2 x} = \sec^2 x – \operatorname{cosec}^2 x$. Therefore $\int (\sec^2 x – \operatorname{cosec}^2 x)\,dx = \tan x + \cot x + C$. The correct option is (A).

24. $\int \frac{e^x(1 + x)}{\cos^2(e^x x)}\,dx$ equals

(A) $-\cot(e^x x) + C$    (B) $\tan(x e^x) + C$    (C) $\tan(e^x) + C$    (D) $\cot(e^x) + C$

Answer: Since $\frac{d}{dx}(x e^x) = e^x + x e^x = e^x(1 + x)$, put $t = x e^x$, so $dt = e^x(1 + x)\,dx$. Then $\int \frac{dt}{\cos^2 t} = \int \sec^2 t\,dt = \tan t + C = \tan(x e^x) + C$. The correct option is (B).

Exercise 7.4

These integrals use the standard forms $\int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\log\left|\frac{x – a}{x + a}\right| + C$, $\int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\log\left|\frac{a + x}{a – x}\right| + C$, $\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$, $\int \frac{dx}{\sqrt{x^2 – a^2}} = \log|x + \sqrt{x^2 – a^2}| + C$, $\int \frac{dx}{\sqrt{a^2 – x^2}} = \sin^{-1}\frac{x}{a} + C$ and $\int \frac{dx}{\sqrt{x^2 + a^2}} = \log|x + \sqrt{x^2 + a^2}| + C$, together with completing the square.

Integrate the functions in Exercises 1 to 23.

1. $\frac{3x^2}{x^6 + 1}$

Answer: Put $t = x^3$, so $dt = 3x^2\,dx$. Then $\int \frac{3x^2}{x^6 + 1}\,dx = \int \frac{dt}{t^2 + 1} = \tan^{-1} t + C = \tan^{-1}(x^3) + C$.

2. $\frac{1}{\sqrt{1 + 4x^2}}$

Answer: Put $t = 2x$, so $dt = 2\,dx$. Then $\int \frac{dx}{\sqrt{1 + 4x^2}} = \frac{1}{2}\int \frac{dt}{\sqrt{1 + t^2}} = \frac{1}{2}\log|t + \sqrt{1 + t^2}| + C = \frac{1}{2}\log|2x + \sqrt{1 + 4x^2}| + C$.

3. $\frac{1}{\sqrt{(2 – x)^2 + 1}}$

Answer: Put $t = x – 2$, so $dt = dx$ and $(2 – x)^2 = t^2$. Then $\int \frac{dx}{\sqrt{t^2 + 1}} = \log|t + \sqrt{t^2 + 1}| + C = \log|x – 2 + \sqrt{x^2 – 4x + 5}| + C$.

4. $\frac{1}{\sqrt{9 – 25x^2}}$

Answer: $9 – 25x^2 = 25\left[\left(\frac{3}{5}\right)^2 – x^2\right]$, so $\sqrt{9 – 25x^2} = 5\sqrt{\left(\frac{3}{5}\right)^2 – x^2}$. Then $\int \frac{dx}{\sqrt{9 – 25x^2}} = \frac{1}{5}\sin^{-1}\frac{x}{3/5} + C = \frac{1}{5}\sin^{-1}\frac{5x}{3} + C$.

5. $\frac{3x}{1 + 2x^4}$

Answer: Put $t = x^2$, so $dt = 2x\,dx$, i.e. $x\,dx = \frac{dt}{2}$. Then $\int \frac{3x}{1 + 2x^4}\,dx = \frac{3}{2}\int \frac{dt}{1 + 2t^2} = \frac{3}{2}\cdot\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\,t) + C$.

$$\int \frac{3x}{1 + 2x^4}\,dx = \frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}\,x^2) + C$$

6. $\frac{x^2}{1 – x^6}$

Answer: Put $t = x^3$, so $dt = 3x^2\,dx$, i.e. $x^2\,dx = \frac{dt}{3}$. Then $\int \frac{x^2}{1 – x^6}\,dx = \frac{1}{3}\int \frac{dt}{1 – t^2} = \frac{1}{3}\cdot\frac{1}{2}\log\left|\frac{1 + t}{1 – t}\right| + C = \frac{1}{6}\log\left|\frac{1 + x^3}{1 – x^3}\right| + C$.

7. $\frac{x – 1}{\sqrt{x^2 – 1}}$

Answer: Split as $\frac{x}{\sqrt{x^2 – 1}} – \frac{1}{\sqrt{x^2 – 1}}$. For the first, $t = x^2 – 1$ gives $\int \frac{x}{\sqrt{x^2 – 1}}\,dx = \sqrt{x^2 – 1}$; the second is a standard form. Therefore

$$\int \frac{x – 1}{\sqrt{x^2 – 1}}\,dx = \sqrt{x^2 – 1} – \log|x + \sqrt{x^2 – 1}| + C$$

8. $\frac{x^2}{\sqrt{x^6 + a^6}}$

Answer: Put $t = x^3$, so $dt = 3x^2\,dx$. Since $x^6 + a^6 = t^2 + (a^3)^2$,

$$\int \frac{x^2}{\sqrt{x^6 + a^6}}\,dx = \frac{1}{3}\int \frac{dt}{\sqrt{t^2 + (a^3)^2}} = \frac{1}{3}\log|x^3 + \sqrt{x^6 + a^6}| + C$$

9. $\frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}$

Answer: Put $t = \tan x$, so $dt = \sec^2 x\,dx$. Then $\int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\,dx = \int \frac{dt}{\sqrt{t^2 + 2^2}} = \log|t + \sqrt{t^2 + 4}| + C = \log|\tan x + \sqrt{\tan^2 x + 4}| + C$.

10. $\frac{1}{\sqrt{x^2 + 2x + 2}}$

Answer: $x^2 + 2x + 2 = (x + 1)^2 + 1$. Put $t = x + 1$. Then $\int \frac{dx}{\sqrt{(x + 1)^2 + 1}} = \log|t + \sqrt{t^2 + 1}| + C = \log|x + 1 + \sqrt{x^2 + 2x + 2}| + C$.

11. $\frac{1}{9x^2 + 6x + 5}$

Answer: $9x^2 + 6x + 5 = (3x + 1)^2 + 4$. Put $t = 3x + 1$, so $dt = 3\,dx$. Then $\int \frac{dx}{(3x + 1)^2 + 4} = \frac{1}{3}\int \frac{dt}{t^2 + 2^2} = \frac{1}{3}\cdot\frac{1}{2}\tan^{-1}\frac{t}{2} + C = \frac{1}{6}\tan^{-1}\frac{3x + 1}{2} + C$.

12. $\frac{1}{\sqrt{7 – 6x – x^2}}$

Answer: $7 – 6x – x^2 = 16 – (x + 3)^2$. Put $t = x + 3$. Then $\int \frac{dx}{\sqrt{16 – (x + 3)^2}} = \sin^{-1}\frac{t}{4} + C = \sin^{-1}\frac{x + 3}{4} + C$.

13. $\frac{1}{\sqrt{(x – 1)(x – 2)}}$

Answer: $(x – 1)(x – 2) = x^2 – 3x + 2 = \left(x – \frac{3}{2}\right)^2 – \left(\frac{1}{2}\right)^2$. Put $t = x – \frac{3}{2}$. Then

$$\int \frac{dx}{\sqrt{(x – 1)(x – 2)}} = \log\left|x – \tfrac{3}{2} + \sqrt{x^2 – 3x + 2}\right| + C$$

14. $\frac{1}{\sqrt{8 + 3x – x^2}}$

Answer: $8 + 3x – x^2 = \frac{41}{4} – \left(x – \frac{3}{2}\right)^2$. Put $t = x – \frac{3}{2}$, with $a = \frac{\sqrt{41}}{2}$. Then

$$\int \frac{dx}{\sqrt{8 + 3x – x^2}} = \sin^{-1}\frac{x – 3/2}{\sqrt{41}/2} + C = \sin^{-1}\frac{2x – 3}{\sqrt{41}} + C$$

15. $\frac{1}{\sqrt{(x – a)(x – b)}}$

Answer: $(x – a)(x – b) = \left(x – \frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$. Put $t = x – \frac{a + b}{2}$. Then

$$\int \frac{dx}{\sqrt{(x – a)(x – b)}} = \log\left|x – \tfrac{a + b}{2} + \sqrt{(x – a)(x – b)}\right| + C$$

16. $\frac{4x + 1}{\sqrt{2x^2 + x – 3}}$

Answer: Note $\frac{d}{dx}(2x^2 + x – 3) = 4x + 1$, exactly the numerator. Put $t = 2x^2 + x – 3$, so $dt = (4x + 1)\,dx$. Then $\int \frac{4x + 1}{\sqrt{2x^2 + x – 3}}\,dx = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} + C = 2\sqrt{2x^2 + x – 3} + C$.

17. $\frac{x + 2}{\sqrt{x^2 – 1}}$

Answer: Write $x + 2 = \frac{1}{2}(2x) + 2$. Then $\int \frac{x + 2}{\sqrt{x^2 – 1}}\,dx = \frac{1}{2}\int \frac{2x}{\sqrt{x^2 – 1}}\,dx + 2\int \frac{dx}{\sqrt{x^2 – 1}} = \sqrt{x^2 – 1} + 2\log|x + \sqrt{x^2 – 1}| + C$.

18. $\frac{5x – 2}{1 + 2x + 3x^2}$

Answer: Write $5x – 2 = A(6x + 2) + B$, where $6x + 2 = \frac{d}{dx}(3x^2 + 2x + 1)$. Comparing, $6A = 5$ so $A = \frac{5}{6}$, and $2A + B = -2$ so $B = -\frac{11}{3}$. Also $3x^2 + 2x + 1 = 3\left[\left(x + \frac{1}{3}\right)^2 + \frac{2}{9}\right]$. Hence

$$\int \frac{5x – 2}{3x^2 + 2x + 1}\,dx = \frac{5}{6}\log|3x^2 + 2x + 1| – \frac{11}{3\sqrt{2}}\tan^{-1}\frac{3x + 1}{\sqrt{2}} + C$$

19. $\frac{6x + 7}{\sqrt{(x – 5)(x – 4)}}$

Answer: Here $(x – 5)(x – 4) = x^2 – 9x + 20$, and $\frac{d}{dx}(x^2 – 9x + 20) = 2x – 9$. Write $6x + 7 = 3(2x – 9) + 34$. Then

$$\int \frac{6x + 7}{\sqrt{x^2 – 9x + 20}}\,dx = 6\sqrt{x^2 – 9x + 20} + 34\log\left|x – \tfrac{9}{2} + \sqrt{x^2 – 9x + 20}\right| + C$$

20. $\frac{x + 2}{\sqrt{4x – x^2}}$

Answer: Here $\frac{d}{dx}(4x – x^2) = 4 – 2x$. Write $x + 2 = -\frac{1}{2}(4 – 2x) + 4$. Also $4x – x^2 = 4 – (x – 2)^2$. Then

$$\int \frac{x + 2}{\sqrt{4x – x^2}}\,dx = -\sqrt{4x – x^2} + 4\sin^{-1}\frac{x – 2}{2} + C$$

21. $\frac{x + 2}{\sqrt{x^2 + 2x + 3}}$

Answer: Here $\frac{d}{dx}(x^2 + 2x + 3) = 2x + 2$. Write $x + 2 = \frac{1}{2}(2x + 2) + 1$. Also $x^2 + 2x + 3 = (x + 1)^2 + 2$. Then

$$\int \frac{x + 2}{\sqrt{x^2 + 2x + 3}}\,dx = \sqrt{x^2 + 2x + 3} + \log|x + 1 + \sqrt{x^2 + 2x + 3}| + C$$

22. $\frac{x + 3}{x^2 – 2x – 5}$

Answer: Here $\frac{d}{dx}(x^2 – 2x – 5) = 2x – 2$. Write $x + 3 = \frac{1}{2}(2x – 2) + 4$. Also $x^2 – 2x – 5 = (x – 1)^2 – 6$. Then

$$\int \frac{x + 3}{x^2 – 2x – 5}\,dx = \frac{1}{2}\log|x^2 – 2x – 5| + \frac{2}{\sqrt{6}}\log\left|\frac{x – 1 – \sqrt{6}}{x – 1 + \sqrt{6}}\right| + C$$

23. $\frac{5x + 3}{\sqrt{x^2 + 4x + 10}}$

Answer: Here $\frac{d}{dx}(x^2 + 4x + 10) = 2x + 4$. Write $5x + 3 = \frac{5}{2}(2x + 4) – 7$. Also $x^2 + 4x + 10 = (x + 2)^2 + 6$. Then

$$\int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}}\,dx = 5\sqrt{x^2 + 4x + 10} – 7\log|x + 2 + \sqrt{x^2 + 4x + 10}| + C$$

Choose the correct answer in Exercises 24 and 25.

24. $\int \frac{dx}{x^2 + 2x + 2}$ equals

(A) $x\tan^{-1}(x + 1) + C$    (B) $\tan^{-1}(x + 1) + C$    (C) $(x + 1)\tan^{-1} x + C$    (D) $\tan^{-1} x + C$

Answer: $x^2 + 2x + 2 = (x + 1)^2 + 1$, so $\int \frac{dx}{(x + 1)^2 + 1} = \tan^{-1}(x + 1) + C$. The correct option is (B).

25. $\int \frac{dx}{\sqrt{9x – 4x^2}}$ equals

(A) $\frac{1}{9}\sin^{-1}\left(\frac{9x – 8}{8}\right) + C$    (B) $\frac{1}{2}\sin^{-1}\left(\frac{8x – 9}{9}\right) + C$    (C) $\frac{1}{3}\sin^{-1}\left(\frac{9x – 8}{8}\right) + C$    (D) $\frac{1}{2}\sin^{-1}\left(\frac{9x – 8}{9}\right) + C$

Answer: $9x – 4x^2 = 4\left[\left(\frac{9}{8}\right)^2 – \left(x – \frac{9}{8}\right)^2\right]$, so $\sqrt{9x – 4x^2} = 2\sqrt{\left(\frac{9}{8}\right)^2 – \left(x – \frac{9}{8}\right)^2}$. Hence $\int \frac{dx}{\sqrt{9x – 4x^2}} = \frac{1}{2}\sin^{-1}\frac{x – 9/8}{9/8} + C = \frac{1}{2}\sin^{-1}\frac{8x – 9}{9} + C$. The correct option is (B).

Exercise 7.5

Integrate the rational functions in Exercises 1 to 21.

1. $\frac{x}{(x+1)(x+2)}$

Answer: Write $\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$, so $x = A(x+2) + B(x+1)$. Putting $x = -1$ gives $A = -1$; putting $x = -2$ gives $B = 2$. Therefore

$$\int \frac{x}{(x+1)(x+2)}\,dx = -\log|x+1| + 2\log|x+2| + C$$

2. $\frac{1}{x^2 – 9}$

Answer: Since $x^2 – 9 = (x-3)(x+3)$, using $\int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|$ with $a = 3$,

$$\int \frac{dx}{x^2 – 9} = \frac{1}{6}\log\left|\frac{x-3}{x+3}\right| + C$$

3. $\frac{3x – 1}{(x-1)(x-2)(x-3)}$

Answer: Take $\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$. By the cover-up rule, $A = \frac{3(1)-1}{(1-2)(1-3)} = 1$, $B = \frac{3(2)-1}{(2-1)(2-3)} = -5$, $C = \frac{3(3)-1}{(3-1)(3-2)} = 4$. Therefore

$$\int \frac{3x-1}{(x-1)(x-2)(x-3)}\,dx = \log|x-1| – 5\log|x-2| + 4\log|x-3| + C$$

4. $\frac{x}{(x-1)(x-2)(x-3)}$

Answer: With $\frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$, the cover-up rule gives $A = \frac{1}{(1-2)(1-3)} = \frac{1}{2}$, $B = \frac{2}{(2-1)(2-3)} = -2$, $C = \frac{3}{(3-1)(3-2)} = \frac{3}{2}$. Therefore

$$\int \frac{x}{(x-1)(x-2)(x-3)}\,dx = \frac{1}{2}\log|x-1| – 2\log|x-2| + \frac{3}{2}\log|x-3| + C$$

5. $\frac{2x}{x^2 + 3x + 2}$

Answer: Here $x^2 + 3x + 2 = (x+1)(x+2)$. Write $\frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$, so $2x = A(x+2) + B(x+1)$. Putting $x = -1$ gives $A = -2$; putting $x = -2$ gives $B = 4$. Therefore

$$\int \frac{2x}{x^2 + 3x + 2}\,dx = -2\log|x+1| + 4\log|x+2| + C$$

6. $\frac{1 – x^2}{x(1 – 2x)}$

Answer: The numerator and denominator both have degree $2$, so divide first. This gives $\frac{1 – x^2}{x(1 – 2x)} = \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1 – 2x)}$. Integrating term by term,

$$\int \frac{1 – x^2}{x(1 – 2x)}\,dx = \frac{x}{2} + \log|x| – \frac{3}{4}\log|1 – 2x| + C$$

7. $\frac{x}{(x^2 + 1)(x – 1)}$

Answer: Write $\frac{x}{(x^2+1)(x-1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x – 1}$, so $x = (Ax+B)(x-1) + C(x^2+1)$. Putting $x = 1$: $1 = 2C$, so $C = \frac{1}{2}$. Comparing coefficients of $x^2$: $A + C = 0 \Rightarrow A = -\frac{1}{2}$; of $x$: $B – A = 1 \Rightarrow B = \frac{1}{2}$. Therefore

$$\int \frac{x}{(x^2+1)(x-1)}\,dx = \frac{1}{2}\log|x-1| – \frac{1}{4}\log(x^2+1) + \frac{1}{2}\tan^{-1}x + C$$

8. $\frac{x}{(x-1)^2 (x+2)}$

Answer: Write $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$, so $x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$. Putting $x = 1$: $1 = 3B \Rightarrow B = \frac{1}{3}$. Putting $x = -2$: $-2 = 9C \Rightarrow C = -\frac{2}{9}$. Comparing coefficients of $x^2$: $A + C = 0 \Rightarrow A = \frac{2}{9}$. Therefore

$$\int \frac{x}{(x-1)^2(x+2)}\,dx = \frac{2}{9}\log|x-1| – \frac{1}{3(x-1)} – \frac{2}{9}\log|x+2| + C$$

9. $\frac{3x + 5}{x^3 – x^2 – x + 1}$

Answer: Factorise the denominator: $x^3 – x^2 – x + 1 = x^2(x-1) – (x-1) = (x-1)(x^2-1) = (x-1)^2(x+1)$. Write $\frac{3x+5}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$, so $3x + 5 = A(x-1)(x+1) + B(x+1) + C(x-1)^2$. Putting $x = 1$: $8 = 2B \Rightarrow B = 4$. Putting $x = -1$: $2 = 4C \Rightarrow C = \frac{1}{2}$. Comparing coefficients of $x^2$: $A + C = 0 \Rightarrow A = -\frac{1}{2}$. Therefore

$$\int \frac{3x+5}{x^3 – x^2 – x + 1}\,dx = -\frac{1}{2}\log|x-1| – \frac{4}{x-1} + \frac{1}{2}\log|x+1| + C$$

10. $\frac{2x – 3}{(x^2 – 1)(2x + 3)}$

Answer: Since $x^2 – 1 = (x-1)(x+1)$, write $\frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3}$. By the cover-up rule, $A = \frac{2(1)-3}{(1+1)(2+3)} = -\frac{1}{10}$, $B = \frac{2(-1)-3}{(-1-1)(-2+3)} = \frac{5}{2}$, and $C = \frac{2(-\frac{3}{2})-3}{(-\frac{3}{2}-1)(-\frac{3}{2}+1)} = -\frac{24}{5}$. Since $\int \frac{C}{2x+3}\,dx = \frac{C}{2}\log|2x+3|$, we get

$$\int \frac{2x-3}{(x^2-1)(2x+3)}\,dx = -\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| – \frac{12}{5}\log|2x+3| + C$$

11. $\frac{5x}{(x+1)(x^2 – 4)}$

Answer: Since $x^2 – 4 = (x-2)(x+2)$, write $\frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2}$. By the cover-up rule, $A = \frac{5(-1)}{(-1-2)(-1+2)} = \frac{5}{3}$, $B = \frac{5(2)}{(2+1)(2+2)} = \frac{5}{6}$, $C = \frac{5(-2)}{(-2+1)(-2-2)} = -\frac{5}{2}$. Therefore

$$\int \frac{5x}{(x+1)(x^2-4)}\,dx = \frac{5}{3}\log|x+1| + \frac{5}{6}\log|x-2| – \frac{5}{2}\log|x+2| + C$$

12. $\frac{x^3 + x + 1}{x^2 – 1}$

Answer: The numerator has higher degree, so divide: $\frac{x^3 + x + 1}{x^2 – 1} = x + \frac{2x + 1}{x^2 – 1}$. Then $\frac{2x+1}{(x-1)(x+1)} = \frac{3/2}{x-1} + \frac{1/2}{x+1}$. Therefore

$$\int \frac{x^3 + x + 1}{x^2 – 1}\,dx = \frac{x^2}{2} + \frac{3}{2}\log|x-1| + \frac{1}{2}\log|x+1| + C$$

13. $\frac{2}{(1 – x)(1 + x^2)}$

Answer: Write $\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx + C}{1+x^2}$, so $2 = A(1+x^2) + (Bx+C)(1-x)$. Putting $x = 1$: $2 = 2A \Rightarrow A = 1$. Comparing coefficients of $x^2$: $A – B = 0 \Rightarrow B = 1$; constant term: $A + C = 2 \Rightarrow C = 1$. Therefore

$$\int \frac{2}{(1-x)(1+x^2)}\,dx = -\log|1-x| + \frac{1}{2}\log(1+x^2) + \tan^{-1}x + C$$

14. $\frac{3x – 1}{(x + 2)^2}$

Answer: Write $3x – 1 = 3(x+2) – 7$, so $\frac{3x-1}{(x+2)^2} = \frac{3}{x+2} – \frac{7}{(x+2)^2}$. Therefore

$$\int \frac{3x-1}{(x+2)^2}\,dx = 3\log|x+2| + \frac{7}{x+2} + C$$

15. $\frac{1}{x^4 – 1}$

Answer: $\frac{1}{x^4 – 1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2}\left(\frac{1}{x^2-1} – \frac{1}{x^2+1}\right)$. Using $\int \frac{dx}{x^2-1} = \frac{1}{2}\log\left|\frac{x-1}{x+1}\right|$ and $\int \frac{dx}{x^2+1} = \tan^{-1}x$,

$$\int \frac{dx}{x^4 – 1} = \frac{1}{4}\log\left|\frac{x-1}{x+1}\right| – \frac{1}{2}\tan^{-1}x + C$$

16. $\frac{1}{x(x^n + 1)}$   [Hint: multiply numerator and denominator by $x^{n-1}$ and put $x^n = t$]

Answer: Multiply numerator and denominator by $x^{n-1}$: $\frac{1}{x(x^n+1)} = \frac{x^{n-1}}{x^n(x^n+1)}$. Put $t = x^n$, so $dt = n\,x^{n-1}\,dx$. Then $\frac{1}{n}\int \frac{dt}{t(t+1)} = \frac{1}{n}\int\left(\frac{1}{t} – \frac{1}{t+1}\right)dt = \frac{1}{n}\log\left|\frac{t}{t+1}\right|$. Therefore

$$\int \frac{dx}{x(x^n + 1)} = \frac{1}{n}\log\left|\frac{x^n}{x^n + 1}\right| + C$$

17. $\frac{\cos x}{(1 – \sin x)(2 – \sin x)}$   [Hint: Put $\sin x = t$]

Answer: Put $t = \sin x$, so $dt = \cos x\,dx$. The integral becomes $\int \frac{dt}{(1-t)(2-t)}$. Since $\frac{1}{(1-t)(2-t)} = \frac{1}{1-t} – \frac{1}{2-t}$, we get $-\log|1-t| + \log|2-t| = \log\left|\frac{2-t}{1-t}\right|$. Therefore

$$\int \frac{\cos x}{(1 – \sin x)(2 – \sin x)}\,dx = \log\left|\frac{2 – \sin x}{1 – \sin x}\right| + C$$

18. $\frac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)}$

Answer: Both numerator and denominator have degree $4$, so $\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)} = \frac{x^4 + 3x^2 + 2}{x^4 + 7x^2 + 12} = 1 – \frac{4x^2 + 10}{(x^2+3)(x^2+4)}$. Treating $x^2$ as the variable, $\frac{4x^2 + 10}{(x^2+3)(x^2+4)} = -\frac{2}{x^2+3} + \frac{6}{x^2+4}$, so the integrand is $1 + \frac{2}{x^2+3} – \frac{6}{x^2+4}$. Therefore

$$\int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx = x + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} – 3\tan^{-1}\frac{x}{2} + C$$

19. $\frac{2x}{(x^2 + 1)(x^2 + 3)}$

Answer: Put $t = x^2$, so $dt = 2x\,dx$. Then $\int \frac{2x\,dx}{(x^2+1)(x^2+3)} = \int \frac{dt}{(t+1)(t+3)} = \frac{1}{2}\int\left(\frac{1}{t+1} – \frac{1}{t+3}\right)dt = \frac{1}{2}\log\left|\frac{t+1}{t+3}\right|$. Therefore

$$\int \frac{2x}{(x^2+1)(x^2+3)}\,dx = \frac{1}{2}\log\frac{x^2 + 1}{x^2 + 3} + C$$

20. $\frac{1}{x(x^4 – 1)}$

Answer: Multiply numerator and denominator by $x^3$: $\frac{1}{x(x^4-1)} = \frac{x^3}{x^4(x^4-1)}$. Put $t = x^4$, so $dt = 4x^3\,dx$. Then $\frac{1}{4}\int \frac{dt}{t(t-1)} = \frac{1}{4}\int\left(\frac{1}{t-1} – \frac{1}{t}\right)dt = \frac{1}{4}\log\left|\frac{t-1}{t}\right|$. Therefore

$$\int \frac{dx}{x(x^4 – 1)} = \frac{1}{4}\log\left|\frac{x^4 – 1}{x^4}\right| + C$$

21. $\frac{1}{e^x – 1}$   [Hint: Put $e^x = t$]

Answer: Put $t = e^x$, so $dt = e^x\,dx = t\,dx$, i.e. $dx = \frac{dt}{t}$. Then $\int \frac{dx}{e^x – 1} = \int \frac{dt}{t(t-1)} = \int\left(\frac{1}{t-1} – \frac{1}{t}\right)dt = \log|t-1| – \log|t|$. Therefore

$$\int \frac{dx}{e^x – 1} = \log|e^x – 1| – x + C$$

Choose the correct answer in each of the Exercises 22 and 23.

22. $\int \frac{x\,dx}{(x-1)(x-2)}$ equals

(A) $\log\left|\frac{(x-1)^2}{x-2}\right| + C$    (B) $\log\left|\frac{(x-2)^2}{x-1}\right| + C$    (C) $\log\left|\left(\frac{x-1}{x-2}\right)^2\right| + C$    (D) $\log|(x-1)(x-2)| + C$

Answer: By partial fractions, $\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}$. Hence the integral $= -\log|x-1| + 2\log|x-2| = \log\left|\frac{(x-2)^2}{x-1}\right| + C$. The correct option is (B).

23. $\int \frac{dx}{x(x^2 + 1)}$ equals

(A) $\log|x| – \frac{1}{2}\log(x^2+1) + C$    (B) $\log|x| + \frac{1}{2}\log(x^2+1) + C$    (C) $-\log|x| + \frac{1}{2}\log(x^2+1) + C$    (D) $\frac{1}{2}\log|x| + \log(x^2+1) + C$

Answer: Since $\frac{1}{x(x^2+1)} = \frac{1}{x} – \frac{x}{x^2+1}$, the integral $= \log|x| – \frac{1}{2}\log(x^2+1) + C$. The correct option is (A).

Exercise 7.6

Integrate the functions in Exercises 1 to 22. (Throughout, integration by parts uses $\int u\,v\,dx = u\int v\,dx – \int\left(u^{\prime}\int v\,dx\right)dx$.)

1. $x \sin x$

Answer: Take $x$ as the first function and $\sin x$ as the second. $\int x\sin x\,dx = x(-\cos x) – \int 1\cdot(-\cos x)\,dx = -x\cos x + \int \cos x\,dx$. Therefore $\int x\sin x\,dx = -x\cos x + \sin x + C$.

2. $x \sin 3x$

Answer: With $x$ first and $\sin 3x$ second, $\int x\sin 3x\,dx = x\left(-\frac{\cos 3x}{3}\right) – \int 1\cdot\left(-\frac{\cos 3x}{3}\right)dx = -\frac{x}{3}\cos 3x + \frac{1}{3}\int\cos 3x\,dx$. Therefore $\int x\sin 3x\,dx = -\frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + C$.

3. $x^2 e^x$

Answer: $\int x^2 e^x\,dx = x^2 e^x – \int 2x\,e^x\,dx = x^2 e^x – 2\left(x e^x – \int e^x\,dx\right)$. Therefore $\int x^2 e^x\,dx = e^x(x^2 – 2x + 2) + C$.

4. $x \log x$

Answer: Take $\log x$ as the first function and $x$ as the second. $\int x\log x\,dx = \log x\cdot\frac{x^2}{2} – \int \frac{1}{x}\cdot\frac{x^2}{2}\,dx = \frac{x^2}{2}\log x – \frac{1}{2}\int x\,dx$. Therefore $\int x\log x\,dx = \frac{x^2}{2}\log x – \frac{x^2}{4} + C$.

5. $x \log 2x$

Answer: Take $\log 2x$ first and $x$ second; note $\frac{d}{dx}\log 2x = \frac{1}{x}$. $\int x\log 2x\,dx = \frac{x^2}{2}\log 2x – \int \frac{1}{x}\cdot\frac{x^2}{2}\,dx$. Therefore $\int x\log 2x\,dx = \frac{x^2}{2}\log 2x – \frac{x^2}{4} + C$.

6. $x^2 \log x$

Answer: Take $\log x$ first and $x^2$ second. $\int x^2\log x\,dx = \log x\cdot\frac{x^3}{3} – \int \frac{1}{x}\cdot\frac{x^3}{3}\,dx = \frac{x^3}{3}\log x – \frac{1}{3}\int x^2\,dx$. Therefore $\int x^2\log x\,dx = \frac{x^3}{3}\log x – \frac{x^3}{9} + C$.

7. $x \sin^{-1} x$

Answer: Take $\sin^{-1}x$ first and $x$ second. $\int x\sin^{-1}x\,dx = \frac{x^2}{2}\sin^{-1}x – \frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}\,dx$. Writing $x^2 = 1 – (1-x^2)$ gives $\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \sin^{-1}x – \frac{x}{2}\sqrt{1-x^2} – \frac{1}{2}\sin^{-1}x = \frac{1}{2}\sin^{-1}x – \frac{x}{2}\sqrt{1-x^2}$. Therefore

$$\int x\sin^{-1}x\,dx = \frac{2x^2 – 1}{4}\sin^{-1}x + \frac{x}{4}\sqrt{1-x^2} + C$$

8. $x \tan^{-1} x$

Answer: Take $\tan^{-1}x$ first and $x$ second. $\int x\tan^{-1}x\,dx = \frac{x^2}{2}\tan^{-1}x – \frac{1}{2}\int \frac{x^2}{1+x^2}\,dx$. Since $\frac{x^2}{1+x^2} = 1 – \frac{1}{1+x^2}$, $\int \frac{x^2}{1+x^2}\,dx = x – \tan^{-1}x$. Therefore

$$\int x\tan^{-1}x\,dx = \frac{x^2 + 1}{2}\tan^{-1}x – \frac{x}{2} + C$$

9. $x \cos^{-1} x$

Answer: Take $\cos^{-1}x$ first and $x$ second. $\int x\cos^{-1}x\,dx = \frac{x^2}{2}\cos^{-1}x + \frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}\,dx$. Using $\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \frac{1}{2}\sin^{-1}x – \frac{x}{2}\sqrt{1-x^2}$ and simplifying,

$$\int x\cos^{-1}x\,dx = \frac{2x^2 – 1}{4}\cos^{-1}x – \frac{x}{4}\sqrt{1-x^2} + C$$

10. $(\sin^{-1} x)^2$

Answer: Take $(\sin^{-1}x)^2$ first and $1$ second. $\int (\sin^{-1}x)^2\,dx = x(\sin^{-1}x)^2 – \int x\cdot\frac{2\sin^{-1}x}{\sqrt{1-x^2}}\,dx$. For the second integral put $\sin^{-1}x$ first and $\frac{x}{\sqrt{1-x^2}}$ second: $\int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}\,\sin^{-1}x + x$. Therefore

$$\int (\sin^{-1}x)^2\,dx = x(\sin^{-1}x)^2 + 2\sqrt{1-x^2}\,\sin^{-1}x – 2x + C$$

11. $\frac{x \cos^{-1} x}{\sqrt{1 – x^2}}$

Answer: Take $\cos^{-1}x$ first and $\frac{x}{\sqrt{1-x^2}}$ second, whose integral is $-\sqrt{1-x^2}$. Then $\int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}\,\cos^{-1}x – \int \left(-\sqrt{1-x^2}\right)\left(\frac{-1}{\sqrt{1-x^2}}\right)dx = -\sqrt{1-x^2}\,\cos^{-1}x – \int dx$. Therefore $\int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}\,\cos^{-1}x – x + C$.

12. $x \sec^2 x$

Answer: Take $x$ first and $\sec^2 x$ second, whose integral is $\tan x$. $\int x\sec^2 x\,dx = x\tan x – \int \tan x\,dx = x\tan x – (-\log|\cos x|)$. Therefore $\int x\sec^2 x\,dx = x\tan x + \log|\cos x| + C$.

13. $\tan^{-1} x$

Answer: Take $\tan^{-1}x$ first and $1$ second. $\int \tan^{-1}x\,dx = x\tan^{-1}x – \int \frac{x}{1+x^2}\,dx$. Therefore $\int \tan^{-1}x\,dx = x\tan^{-1}x – \frac{1}{2}\log(1+x^2) + C$.

14. $x (\log x)^2$

Answer: Take $(\log x)^2$ first and $x$ second. $\int x(\log x)^2\,dx = \frac{x^2}{2}(\log x)^2 – \int \frac{x^2}{2}\cdot\frac{2\log x}{x}\,dx = \frac{x^2}{2}(\log x)^2 – \int x\log x\,dx$. Using Exercise 4, $\int x\log x\,dx = \frac{x^2}{2}\log x – \frac{x^2}{4}$. Therefore

$$\int x(\log x)^2\,dx = \frac{x^2}{2}(\log x)^2 – \frac{x^2}{2}\log x + \frac{x^2}{4} + C$$

15. $(x^2 + 1) \log x$

Answer: Split as $\int x^2\log x\,dx + \int \log x\,dx$. From Exercise 6, $\int x^2\log x\,dx = \frac{x^3}{3}\log x – \frac{x^3}{9}$, and $\int \log x\,dx = x\log x – x$. Therefore

$$\int (x^2 + 1)\log x\,dx = \left(\frac{x^3}{3} + x\right)\log x – \frac{x^3}{9} – x + C$$

16. $e^x (\sin x + \cos x)$

Answer: This is of the form $e^x[f(x) + f^{\prime}(x)]$ with $f(x) = \sin x$ and $f^{\prime}(x) = \cos x$. Using $\int e^x[f(x) + f^{\prime}(x)]\,dx = e^x f(x) + C$, we get $\int e^x(\sin x + \cos x)\,dx = e^x \sin x + C$.

17. $\frac{x e^x}{(1 + x)^2}$

Answer: Write $\frac{x}{(1+x)^2} = \frac{(1+x) – 1}{(1+x)^2} = \frac{1}{1+x} – \frac{1}{(1+x)^2}$. With $f(x) = \frac{1}{1+x}$, $f^{\prime}(x) = -\frac{1}{(1+x)^2}$, so the integrand is $e^x[f(x) + f^{\prime}(x)]$. Therefore $\int \frac{x e^x}{(1+x)^2}\,dx = \frac{e^x}{1+x} + C$.

18. $e^x \left(\frac{1 + \sin x}{1 + \cos x}\right)$

Answer: Using $1 + \cos x = 2\cos^2\frac{x}{2}$ and $1 + \sin x = 1 + 2\sin\frac{x}{2}\cos\frac{x}{2}$, we get $\frac{1+\sin x}{1+\cos x} = \frac{1}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2}$. With $f(x) = \tan\frac{x}{2}$, $f^{\prime}(x) = \frac{1}{2}\sec^2\frac{x}{2}$, so the integrand is $e^x[f(x) + f^{\prime}(x)]$. Therefore $\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx = e^x \tan\frac{x}{2} + C$.

19. $e^x \left(\frac{1}{x} – \frac{1}{x^2}\right)$

Answer: With $f(x) = \frac{1}{x}$, $f^{\prime}(x) = -\frac{1}{x^2}$, the integrand is $e^x[f(x) + f^{\prime}(x)]$. Therefore $\int e^x\left(\frac{1}{x} – \frac{1}{x^2}\right)dx = \frac{e^x}{x} + C$.

20. $\frac{(x – 3) e^x}{(x – 1)^3}$

Answer: Write $\frac{x-3}{(x-1)^3} = \frac{(x-1) – 2}{(x-1)^3} = \frac{1}{(x-1)^2} – \frac{2}{(x-1)^3}$. With $f(x) = \frac{1}{(x-1)^2}$, $f^{\prime}(x) = -\frac{2}{(x-1)^3}$, so the integrand is $e^x[f(x) + f^{\prime}(x)]$. Therefore $\int \frac{(x-3)e^x}{(x-1)^3}\,dx = \frac{e^x}{(x-1)^2} + C$.

21. $e^{2x} \sin x$

Answer: Let $I = \int e^{2x}\sin x\,dx$. Integrating by parts twice, $I = e^{2x}(-\cos x) + 2\int e^{2x}\cos x\,dx$ and $\int e^{2x}\cos x\,dx = e^{2x}\sin x – 2I$. Substituting gives $I = -e^{2x}\cos x + 2e^{2x}\sin x – 4I$, so $5I = e^{2x}(2\sin x – \cos x)$. Therefore

$$\int e^{2x}\sin x\,dx = \frac{e^{2x}}{5}(2\sin x – \cos x) + C$$

22. $\sin^{-1}\left(\frac{2x}{1 + x^2}\right)$

Answer: Put $x = \tan\theta$; then $\frac{2x}{1+x^2} = \sin 2\theta$, so $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\theta = 2\tan^{-1}x$. Hence $\int \sin^{-1}\left(\frac{2x}{1+x^2}\right)dx = 2\int \tan^{-1}x\,dx = 2\left(x\tan^{-1}x – \frac{1}{2}\log(1+x^2)\right)$. Therefore

$$\int \sin^{-1}\left(\frac{2x}{1 + x^2}\right)dx = 2x\tan^{-1}x – \log(1 + x^2) + C$$

Choose the correct answer in Exercises 23 and 24.

23. $\int x^2 e^{x^3}\,dx$ equals

(A) $\frac{1}{3}e^{x^3} + C$    (B) $\frac{1}{3}e^{x^2} + C$    (C) $\frac{1}{2}e^{x^3} + C$    (D) $\frac{1}{2}e^{x^2} + C$

Answer: Put $t = x^3$, so $dt = 3x^2\,dx$. Then $\int x^2 e^{x^3}\,dx = \frac{1}{3}\int e^t\,dt = \frac{1}{3}e^{x^3} + C$. The correct option is (A).

24. $\int e^x \sec x\,(1 + \tan x)\,dx$ equals

(A) $e^x \cos x + C$    (B) $e^x \sec x + C$    (C) $e^x \sin x + C$    (D) $e^x \tan x + C$

Answer: $e^x\sec x(1 + \tan x) = e^x(\sec x + \sec x\tan x)$. With $f(x) = \sec x$, $f^{\prime}(x) = \sec x\tan x$, this is $e^x[f(x) + f^{\prime}(x)]$, so the integral $= e^x\sec x + C$. The correct option is (B).

Exercise 7.7

Integrate the functions in Exercises 1 to 9. These use the standard results $\int\sqrt{a^2 – x^2}\,dx = \frac{x}{2}\sqrt{a^2 – x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$, $\int\sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C$, and $\int\sqrt{x^2 – a^2}\,dx = \frac{x}{2}\sqrt{x^2 – a^2} – \frac{a^2}{2}\log\left|x + \sqrt{x^2 – a^2}\right| + C$, with the square completed first where needed.

1. $\sqrt{4 – x^2}$

Answer: Here $a = 2$. Therefore

$$\int \sqrt{4 – x^2}\,dx = \frac{x}{2}\sqrt{4 – x^2} + 2\sin^{-1}\frac{x}{2} + C$$

2. $\sqrt{1 – 4x^2}$

Answer: Write $\sqrt{1 – 4x^2} = 2\sqrt{\frac{1}{4} – x^2}$, so $a = \frac{1}{2}$. Then $\int \sqrt{1 – 4x^2}\,dx = 2\left[\frac{x}{2}\sqrt{\frac{1}{4} – x^2} + \frac{1}{8}\sin^{-1}(2x)\right]$. Therefore

$$\int \sqrt{1 – 4x^2}\,dx = \frac{x}{2}\sqrt{1 – 4x^2} + \frac{1}{4}\sin^{-1}(2x) + C$$

3. $\sqrt{x^2 + 4x + 6}$

Answer: Complete the square: $x^2 + 4x + 6 = (x+2)^2 + 2$, so $a^2 = 2$. Therefore

$$\int \sqrt{x^2 + 4x + 6}\,dx = \frac{x+2}{2}\sqrt{x^2 + 4x + 6} + \log\left|x + 2 + \sqrt{x^2 + 4x + 6}\right| + C$$

4. $\sqrt{x^2 + 4x + 1}$

Answer: $x^2 + 4x + 1 = (x+2)^2 – 3$, so $a^2 = 3$ and the form is $\sqrt{u^2 – a^2}$. Therefore

$$\int \sqrt{x^2 + 4x + 1}\,dx = \frac{x+2}{2}\sqrt{x^2 + 4x + 1} – \frac{3}{2}\log\left|x + 2 + \sqrt{x^2 + 4x + 1}\right| + C$$

5. $\sqrt{1 – 4x – x^2}$

Answer: $1 – 4x – x^2 = 5 – (x+2)^2$, so $a^2 = 5$ and the form is $\sqrt{a^2 – u^2}$. Therefore

$$\int \sqrt{1 – 4x – x^2}\,dx = \frac{x+2}{2}\sqrt{1 – 4x – x^2} + \frac{5}{2}\sin^{-1}\frac{x+2}{\sqrt{5}} + C$$

6. $\sqrt{x^2 + 4x – 5}$

Answer: $x^2 + 4x – 5 = (x+2)^2 – 9$, so $a^2 = 9$ and the form is $\sqrt{u^2 – a^2}$. Therefore

$$\int \sqrt{x^2 + 4x – 5}\,dx = \frac{x+2}{2}\sqrt{x^2 + 4x – 5} – \frac{9}{2}\log\left|x + 2 + \sqrt{x^2 + 4x – 5}\right| + C$$

7. $\sqrt{1 + 3x – x^2}$

Answer: $1 + 3x – x^2 = \frac{13}{4} – \left(x – \frac{3}{2}\right)^2$, so $a^2 = \frac{13}{4}$ and the form is $\sqrt{a^2 – u^2}$ with $u = x – \frac{3}{2}$. Therefore

$$\int \sqrt{1 + 3x – x^2}\,dx = \frac{2x – 3}{4}\sqrt{1 + 3x – x^2} + \frac{13}{8}\sin^{-1}\frac{2x – 3}{\sqrt{13}} + C$$

8. $\sqrt{x^2 + 3x}$

Answer: $x^2 + 3x = \left(x + \frac{3}{2}\right)^2 – \frac{9}{4}$, so $a^2 = \frac{9}{4}$ and the form is $\sqrt{u^2 – a^2}$ with $u = x + \frac{3}{2}$. Therefore

$$\int \sqrt{x^2 + 3x}\,dx = \frac{2x + 3}{4}\sqrt{x^2 + 3x} – \frac{9}{8}\log\left|x + \frac{3}{2} + \sqrt{x^2 + 3x}\right| + C$$

9. $\sqrt{1 + \frac{x^2}{9}}$

Answer: $\sqrt{1 + \frac{x^2}{9}} = \frac{1}{3}\sqrt{x^2 + 9}$, so $a^2 = 9$. Then $\int \sqrt{1 + \frac{x^2}{9}}\,dx = \frac{1}{3}\left[\frac{x}{2}\sqrt{x^2 + 9} + \frac{9}{2}\log\left|x + \sqrt{x^2 + 9}\right|\right]$. Therefore

$$\int \sqrt{1 + \frac{x^2}{9}}\,dx = \frac{x}{2}\sqrt{1 + \frac{x^2}{9}} + \frac{3}{2}\log\left|x + \sqrt{x^2 + 9}\right| + C$$

Choose the correct answer in Exercises 10 to 11.

10. $\int \sqrt{1 + x^2}\,dx$ is equal to

(A) $\frac{x}{2}\sqrt{1 + x^2} + \frac{1}{2}\log\left|x + \sqrt{1 + x^2}\right| + C$    (B) $\frac{2}{3}(1 + x^2)^{3/2} + C$    (C) $\frac{2}{3}x(1 + x^2)^{3/2} + C$    (D) $\frac{x^2}{2}\sqrt{1 + x^2} + \frac{1}{2}x^2\log\left|x + \sqrt{1 + x^2}\right| + C$

Answer: With $a = 1$ in $\int\sqrt{x^2 + a^2}\,dx$, the integral $= \frac{x}{2}\sqrt{1 + x^2} + \frac{1}{2}\log\left|x + \sqrt{1 + x^2}\right| + C$. The correct option is (A).

11. $\int \sqrt{x^2 – 8x + 7}\,dx$ is equal to

(A) $\frac{1}{2}(x-4)\sqrt{x^2 – 8x + 7} + 9\log\left|x – 4 + \sqrt{x^2 – 8x + 7}\right| + C$    (B) $\frac{1}{2}(x+4)\sqrt{x^2 – 8x + 7} + 9\log\left|x + 4 + \sqrt{x^2 – 8x + 7}\right| + C$    (C) $\frac{1}{2}(x-4)\sqrt{x^2 – 8x + 7} – 3\sqrt{2}\log\left|x – 4 + \sqrt{x^2 – 8x + 7}\right| + C$    (D) $\frac{1}{2}(x-4)\sqrt{x^2 – 8x + 7} – \frac{9}{2}\log\left|x – 4 + \sqrt{x^2 – 8x + 7}\right| + C$

Answer: $x^2 – 8x + 7 = (x-4)^2 – 9$, so $a^2 = 9$ and the form is $\sqrt{u^2 – a^2}$ with $u = x – 4$. Hence the integral $= \frac{x-4}{2}\sqrt{x^2 – 8x + 7} – \frac{9}{2}\log\left|x – 4 + \sqrt{x^2 – 8x + 7}\right| + C$. The correct option is (D).

Exercise 7.8

Evaluate the definite integrals in Exercises 1 to 20.

1. $\int_{-1}^{1} (x + 1)\,dx$

Answer: $\int_{-1}^{1}(x+1)\,dx = \left[\frac{x^2}{2} + x\right]_{-1}^{1} = \left(\frac{1}{2} + 1\right) – \left(\frac{1}{2} – 1\right) = \frac{3}{2} + \frac{1}{2} = 2$.

2. $\int_{2}^{3} \frac{1}{x}\,dx$

Answer: $\int_{2}^{3}\frac{1}{x}\,dx = \left[\log x\right]_{2}^{3} = \log 3 – \log 2 = \log\frac{3}{2}$.

3. $\int_{1}^{2} (4x^3 – 5x^2 + 6x + 9)\,dx$

Answer: $\int_{1}^{2}(4x^3 – 5x^2 + 6x + 9)\,dx = \left[x^4 – \frac{5x^3}{3} + 3x^2 + 9x\right]_{1}^{2}$. At $x=2$ this is $\frac{98}{3}$ and at $x=1$ it is $\frac{34}{3}$, so the value is $\frac{98}{3} – \frac{34}{3} = \frac{64}{3}$.

4. $\int_{0}^{\pi/4} \sin 2x\,dx$

Answer: $\int_{0}^{\pi/4}\sin 2x\,dx = \left[-\frac{1}{2}\cos 2x\right]_{0}^{\pi/4} = -\frac{1}{2}\left(\cos\frac{\pi}{2} – \cos 0\right) = -\frac{1}{2}(0 – 1) = \frac{1}{2}$.

5. $\int_{0}^{\pi/2} \cos 2x\,dx$

Answer: $\int_{0}^{\pi/2}\cos 2x\,dx = \left[\frac{1}{2}\sin 2x\right]_{0}^{\pi/2} = \frac{1}{2}(\sin\pi – \sin 0) = 0$.

6. $\int_{4}^{5} e^x\,dx$

Answer: $\int_{4}^{5} e^x\,dx = \left[e^x\right]_{4}^{5} = e^5 – e^4$.

7. $\int_{0}^{\pi/4} \tan x\,dx$

Answer: $\int_{0}^{\pi/4}\tan x\,dx = \left[-\log|\cos x|\right]_{0}^{\pi/4} = -\log\frac{1}{\sqrt{2}} + \log 1 = \frac{1}{2}\log 2$.

8. $\int_{\pi/6}^{\pi/4} \operatorname{cosec} x\,dx$

Answer: Since $\int \operatorname{cosec} x\,dx = \log\left|\tan\frac{x}{2}\right|$, and $\tan\frac{\pi}{8} = \sqrt{2} – 1$, $\tan\frac{\pi}{12} = 2 – \sqrt{3}$,

$$\int_{\pi/6}^{\pi/4} \operatorname{cosec} x\,dx = \log(\sqrt{2} – 1) – \log(2 – \sqrt{3}) = \log\left[(\sqrt{2} – 1)(2 + \sqrt{3})\right]$$

9. $\int_{0}^{1} \frac{dx}{\sqrt{1 – x^2}}$

Answer: $\int_{0}^{1}\frac{dx}{\sqrt{1 – x^2}} = \left[\sin^{-1}x\right]_{0}^{1} = \sin^{-1}1 – \sin^{-1}0 = \frac{\pi}{2}$.

10. $\int_{0}^{1} \frac{dx}{1 + x^2}$

Answer: $\int_{0}^{1}\frac{dx}{1 + x^2} = \left[\tan^{-1}x\right]_{0}^{1} = \tan^{-1}1 – \tan^{-1}0 = \frac{\pi}{4}$.

11. $\int_{2}^{3} \frac{dx}{x^2 – 1}$

Answer: $\int_{2}^{3}\frac{dx}{x^2 – 1} = \left[\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|\right]_{2}^{3} = \frac{1}{2}\left(\log\frac{2}{4} – \log\frac{1}{3}\right) = \frac{1}{2}\log\frac{3}{2}$.

12. $\int_{0}^{\pi/2} \cos^2 x\,dx$

Answer: Using $\cos^2 x = \frac{1 + \cos 2x}{2}$, $\int_{0}^{\pi/2}\cos^2 x\,dx = \left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_{0}^{\pi/2} = \frac{\pi}{4}$.

13. $\int_{2}^{3} \frac{x\,dx}{x^2 + 1}$

Answer: $\int_{2}^{3}\frac{x}{x^2 + 1}\,dx = \left[\frac{1}{2}\log(x^2 + 1)\right]_{2}^{3} = \frac{1}{2}(\log 10 – \log 5) = \frac{1}{2}\log 2$.

14. $\int_{0}^{1} \frac{2x + 3}{5x^2 + 1}\,dx$

Answer: Split as $\frac{2x}{5x^2 + 1} + \frac{3}{5x^2 + 1}$. Then $\int \frac{2x}{5x^2 + 1}\,dx = \frac{1}{5}\log(5x^2 + 1)$ and $\int \frac{3}{5x^2 + 1}\,dx = \frac{3}{\sqrt{5}}\tan^{-1}(\sqrt{5}\,x)$. Therefore

$$\int_{0}^{1} \frac{2x + 3}{5x^2 + 1}\,dx = \left[\frac{1}{5}\log(5x^2 + 1) + \frac{3}{\sqrt{5}}\tan^{-1}(\sqrt{5}\,x)\right]_{0}^{1} = \frac{1}{5}\log 6 + \frac{3}{\sqrt{5}}\tan^{-1}\sqrt{5}$$

15. $\int_{0}^{1} x e^{x^2}\,dx$

Answer: Put $t = x^2$, so $dt = 2x\,dx$; as $x$ goes $0 \to 1$, $t$ goes $0 \to 1$. Then $\int_{0}^{1} x e^{x^2}\,dx = \frac{1}{2}\int_{0}^{1} e^t\,dt = \frac{1}{2}\left[e^t\right]_{0}^{1} = \frac{1}{2}(e – 1)$.

16. $\int_{1}^{2} \frac{5x^2}{x^2 + 4x + 3}\,dx$

Answer: Since the degrees are equal, $\frac{5x^2}{x^2 + 4x + 3} = 5 – \frac{20x + 15}{(x+1)(x+3)} = 5 + \frac{5/2}{x+1} – \frac{45/2}{x+3}$. Therefore

$$\int_{1}^{2} \frac{5x^2}{x^2 + 4x + 3}\,dx = \left[5x + \frac{5}{2}\log|x+1| – \frac{45}{2}\log|x+3|\right]_{1}^{2} = 5 + \frac{5}{2}\log\frac{3}{2} – \frac{45}{2}\log\frac{5}{4}$$

17. $\int_{0}^{\pi/4} (2\sec^2 x + x^3 + 2)\,dx$

Answer: $\int_{0}^{\pi/4}(2\sec^2 x + x^3 + 2)\,dx = \left[2\tan x + \frac{x^4}{4} + 2x\right]_{0}^{\pi/4} = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2}$.

18. $\int_{0}^{\pi} \left(\sin^2\frac{x}{2} – \cos^2\frac{x}{2}\right)dx$

Answer: Since $\sin^2\frac{x}{2} – \cos^2\frac{x}{2} = -\cos x$, $\int_{0}^{\pi}\left(\sin^2\frac{x}{2} – \cos^2\frac{x}{2}\right)dx = \left[-\sin x\right]_{0}^{\pi} = 0$.

19. $\int_{0}^{2} \frac{6x + 3}{x^2 + 4}\,dx$

Answer: Split as $\frac{6x}{x^2 + 4} + \frac{3}{x^2 + 4}$. Then $\int \frac{6x}{x^2 + 4}\,dx = 3\log(x^2 + 4)$ and $\int \frac{3}{x^2 + 4}\,dx = \frac{3}{2}\tan^{-1}\frac{x}{2}$. Therefore

$$\int_{0}^{2} \frac{6x + 3}{x^2 + 4}\,dx = \left[3\log(x^2 + 4) + \frac{3}{2}\tan^{-1}\frac{x}{2}\right]_{0}^{2} = 3\log 2 + \frac{3\pi}{8}$$

20. $\int_{0}^{1} \left(x e^x + \sin\frac{\pi x}{4}\right)dx$

Answer: By parts, $\int_{0}^{1} x e^x\,dx = \left[e^x(x – 1)\right]_{0}^{1} = 0 – (-1) = 1$. Also $\int_{0}^{1}\sin\frac{\pi x}{4}\,dx = \left[-\frac{4}{\pi}\cos\frac{\pi x}{4}\right]_{0}^{1} = \frac{4}{\pi}\left(1 – \frac{1}{\sqrt{2}}\right)$. Adding,

$$\int_{0}^{1} \left(x e^x + \sin\frac{\pi x}{4}\right)dx = 1 + \frac{4 – 2\sqrt{2}}{\pi}$$

Choose the correct answer in Exercises 21 and 22.

21. $\int_{1}^{\sqrt{3}} \frac{dx}{1 + x^2}$ equals

(A) $\frac{\pi}{3}$    (B) $\frac{2\pi}{3}$    (C) $\frac{\pi}{6}$    (D) $\frac{\pi}{12}$

Answer: $\int_{1}^{\sqrt{3}}\frac{dx}{1 + x^2} = \left[\tan^{-1}x\right]_{1}^{\sqrt{3}} = \tan^{-1}\sqrt{3} – \tan^{-1}1 = \frac{\pi}{3} – \frac{\pi}{4} = \frac{\pi}{12}$. The correct option is (D).

22. $\int_{0}^{2/3} \frac{dx}{4 + 9x^2}$ equals

(A) $\frac{\pi}{6}$    (B) $\frac{\pi}{12}$    (C) $\frac{\pi}{24}$    (D) $\frac{\pi}{4}$

Answer: $\int_{0}^{2/3}\frac{dx}{4 + 9x^2} = \left[\frac{1}{6}\tan^{-1}\frac{3x}{2}\right]_{0}^{2/3} = \frac{1}{6}\tan^{-1}1 = \frac{1}{6}\cdot\frac{\pi}{4} = \frac{\pi}{24}$. The correct option is (C).

Exercise 7.9

Evaluate the integrals in Exercises 1 to 8 using substitution.

1. $\int_0^1 \dfrac{x}{x^2 + 1}\,dx$

Answer: Put $t = x^2 + 1$, so $dt = 2x\,dx$. When $x = 0$, $t = 1$ and when $x = 1$, $t = 2$.

$$\int_0^1 \frac{x}{x^2 + 1}\,dx = \frac{1}{2}\int_1^2 \frac{dt}{t} = \frac{1}{2}\big[\log t\big]_1^2 = \frac{1}{2}\log 2$$

2. $\int_0^{\pi/2} \sqrt{\sin\phi}\,\cos^5\phi\,d\phi$

Answer: Write $\cos^5\phi = \cos^4\phi\cdot\cos\phi = (1 – \sin^2\phi)^2\cos\phi$. Put $t = \sin\phi$, so $dt = \cos\phi\,d\phi$; limits $0 \to 0$ and $\tfrac{\pi}{2} \to 1$.

$$I = \int_0^1 \sqrt{t}\,(1 – t^2)^2\,dt = \int_0^1 \left(t^{1/2} – 2t^{5/2} + t^{9/2}\right)dt$$

$$= \left[\frac{2}{3}t^{3/2} – \frac{4}{7}t^{7/2} + \frac{2}{11}t^{11/2}\right]_0^1 = \frac{2}{3} – \frac{4}{7} + \frac{2}{11} = \frac{154 – 132 + 42}{231} = \frac{64}{231}$$

3. $\int_0^1 \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)dx$

Answer: For $x \in [0, 1]$ we have $\sin^{-1}\dfrac{2x}{1 + x^2} = 2\tan^{-1}x$. Hence, integrating by parts with $\int\tan^{-1}x\,dx = x\tan^{-1}x – \tfrac{1}{2}\log(1 + x^2)$,

$$I = 2\int_0^1 \tan^{-1}x\,dx = 2\left[x\tan^{-1}x – \frac{1}{2}\log(1 + x^2)\right]_0^1 = 2\left(\frac{\pi}{4} – \frac{1}{2}\log 2\right) = \frac{\pi}{2} – \log 2$$

4. $\int_0^2 x\sqrt{x + 2}\,dx$ (Put $x + 2 = t^2$)

Answer: With $x + 2 = t^2$, $x = t^2 – 2$ and $dx = 2t\,dt$. When $x = 0$, $t = \sqrt{2}$ and when $x = 2$, $t = 2$.

$$I = \int_{\sqrt{2}}^{2} (t^2 – 2)\,t\,(2t)\,dt = 2\int_{\sqrt{2}}^{2}(t^4 – 2t^2)\,dt = 2\left[\frac{t^5}{5} – \frac{2t^3}{3}\right]_{\sqrt{2}}^{2}$$

At $t = 2$ the bracket is $\tfrac{32}{5} – \tfrac{16}{3} = \tfrac{16}{15}$; at $t = \sqrt{2}$ it is $\tfrac{4\sqrt{2}}{5} – \tfrac{4\sqrt{2}}{3} = -\tfrac{8\sqrt{2}}{15}$. Therefore

$$I = 2\left(\frac{16}{15} + \frac{8\sqrt{2}}{15}\right) = \frac{16\left(2 + \sqrt{2}\right)}{15}$$

5. $\int_0^{\pi/2} \dfrac{\sin x}{1 + \cos^2 x}\,dx$

Answer: Put $t = \cos x$, so $dt = -\sin x\,dx$; limits $0 \to 1$ and $\tfrac{\pi}{2} \to 0$.

$$I = \int_1^0 \frac{-dt}{1 + t^2} = \int_0^1 \frac{dt}{1 + t^2} = \big[\tan^{-1}t\big]_0^1 = \frac{\pi}{4}$$

6. $\int_0^2 \dfrac{dx}{x + 4 – x^2}$

Answer: Complete the square: $x + 4 – x^2 = \tfrac{17}{4} – \left(x – \tfrac{1}{2}\right)^2$. With $a = \tfrac{\sqrt{17}}{2}$ and $u = x – \tfrac{1}{2}$,

$$I = \int_0^2 \frac{dx}{\frac{17}{4} – \left(x – \frac{1}{2}\right)^2} = \frac{1}{\sqrt{17}}\left[\log\left|\frac{\sqrt{17} + 2x – 1}{\sqrt{17} – 2x + 1}\right|\right]_0^2$$

$$= \frac{1}{\sqrt{17}}\log\!\left[\frac{(\sqrt{17} + 3)(\sqrt{17} + 1)}{(\sqrt{17} – 3)(\sqrt{17} – 1)}\right] = \frac{1}{\sqrt{17}}\log\!\left(\frac{5 + \sqrt{17}}{5 – \sqrt{17}}\right)$$

7. $\int_{-1}^{1} \dfrac{dx}{x^2 + 2x + 5}$

Answer: Since $x^2 + 2x + 5 = (x + 1)^2 + 2^2$,

$$I = \int_{-1}^{1} \frac{dx}{(x + 1)^2 + 2^2} = \frac{1}{2}\left[\tan^{-1}\frac{x + 1}{2}\right]_{-1}^{1} = \frac{1}{2}\left(\tan^{-1}1 – 0\right) = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$$

8. $\int_1^2 \left(\dfrac{1}{x} – \dfrac{1}{2x^2}\right)e^{2x}\,dx$

Answer: Observe that $\dfrac{d}{dx}\!\left(\dfrac{e^{2x}}{2x}\right) = \dfrac{2e^{2x}\cdot 2x – e^{2x}\cdot 2}{4x^2} = \left(\dfrac{1}{x} – \dfrac{1}{2x^2}\right)e^{2x}$. Hence the integrand is an exact derivative.

$$I = \left[\frac{e^{2x}}{2x}\right]_1^2 = \frac{e^4}{4} – \frac{e^2}{2}$$

Choose the correct answer in Exercises 9 and 10.

9. The value of the integral $\int_{1/3}^{1} \dfrac{(x – x^3)^{1/3}}{x^4}\,dx$ is

(A) 6   (B) 0   (C) 3   (D) 4

Answer: (A) 6. Take $x^3$ out of the cube root: $(x – x^3)^{1/3} = x\left(\tfrac{1}{x^2} – 1\right)^{1/3}$, so the integrand becomes $\dfrac{1}{x^3}\left(\dfrac{1}{x^2} – 1\right)^{1/3}$. Put $t = \dfrac{1}{x^2} – 1$, giving $dt = -\dfrac{2}{x^3}\,dx$; limits $x = \tfrac{1}{3} \to t = 8$ and $x = 1 \to t = 0$.

$$I = \frac{1}{2}\int_0^8 t^{1/3}\,dt = \frac{1}{2}\cdot\frac{3}{4}\big[t^{4/3}\big]_0^8 = \frac{3}{8}\cdot 8^{4/3} = \frac{3}{8}\cdot 16 = 6$$

10. If $f(x) = \int_0^x t\sin t\,dt$, then $f^{\prime}(x)$ is

(A) $\cos x + x\sin x$   (B) $x\sin x$   (C) $x\cos x$   (D) $\sin x + x\cos x$

Answer: (B) $x\sin x$. By the first fundamental theorem of calculus, if $f(x) = \int_0^x g(t)\,dt$ then $f^{\prime}(x) = g(x)$. Here $g(t) = t\sin t$, so $f^{\prime}(x) = x\sin x$.

Exercise 7.10

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. Throughout we use $P_4:\ \int_0^a f(x)\,dx = \int_0^a f(a – x)\,dx$ and $P_3:\ \int_a^b f(x)\,dx = \int_a^b f(a + b – x)\,dx$.

1. $\int_0^{\pi/2} \cos^2 x\,dx$

Answer: By $P_4$, $I = \int_0^{\pi/2}\cos^2\!\left(\tfrac{\pi}{2} – x\right)dx = \int_0^{\pi/2}\sin^2 x\,dx$. Adding the two forms,

$$2I = \int_0^{\pi/2}\left(\cos^2 x + \sin^2 x\right)dx = \int_0^{\pi/2} dx = \frac{\pi}{2} \quad\Rightarrow\quad I = \frac{\pi}{4}$$

2. $\int_0^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx$

Answer: By $P_4$, $I = \int_0^{\pi/2}\dfrac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx$. Adding,

$$2I = \int_0^{\pi/2}\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx = \int_0^{\pi/2} dx = \frac{\pi}{2} \quad\Rightarrow\quad I = \frac{\pi}{4}$$

3. $\int_0^{\pi/2} \dfrac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x}\,dx$

Answer: Exactly as in Q2, applying $P_4$ swaps $\sin$ and $\cos$; adding the two forms gives $2I = \int_0^{\pi/2}dx = \dfrac{\pi}{2}$, so $I = \dfrac{\pi}{4}$.

4. $\int_0^{\pi/2} \dfrac{\cos^5 x}{\sin^5 x + \cos^5 x}\,dx$

Answer: By $P_4$, $I = \int_0^{\pi/2}\dfrac{\sin^5 x}{\cos^5 x + \sin^5 x}\,dx$. Adding gives $2I = \int_0^{\pi/2}dx = \dfrac{\pi}{2}$, hence $I = \dfrac{\pi}{4}$.

5. $\int_{-5}^{5} |x + 2|\,dx$

Answer: Here $|x + 2| = -(x + 2)$ for $x < -2$ and $|x + 2| = x + 2$ for $x \ge -2$, so split at $x = -2$.

$$I = \int_{-5}^{-2} -(x + 2)\,dx + \int_{-2}^{5} (x + 2)\,dx = \frac{9}{2} + \frac{49}{2} = 29$$

6. $\int_2^8 |x – 5|\,dx$

Answer: Split at $x = 5$, where $|x – 5|$ changes sign.

$$I = \int_2^5 (5 – x)\,dx + \int_5^8 (x – 5)\,dx = \frac{9}{2} + \frac{9}{2} = 9$$

7. $\int_0^1 x(1 – x)^n\,dx$

Answer: By $P_4$ (with $a = 1$), $I = \int_0^1 (1 – x)\big(1 – (1 – x)\big)^n dx = \int_0^1 (1 – x)x^n\,dx$.

$$I = \int_0^1 \left(x^n – x^{n+1}\right)dx = \frac{1}{n + 1} – \frac{1}{n + 2} = \frac{1}{(n + 1)(n + 2)}$$

8. $\int_0^{\pi/4} \log(1 + \tan x)\,dx$

Answer: By $P_4$ (with $a = \tfrac{\pi}{4}$), replace $x$ by $\tfrac{\pi}{4} – x$. Since $1 + \tan\!\left(\tfrac{\pi}{4} – x\right) = \dfrac{2}{1 + \tan x}$,

$$I = \int_0^{\pi/4}\big[\log 2 – \log(1 + \tan x)\big]dx = \frac{\pi}{4}\log 2 – I \quad\Rightarrow\quad I = \frac{\pi}{8}\log 2$$

9. $\int_0^2 x\sqrt{2 – x}\,dx$

Answer: By $P_4$ (with $a = 2$), $I = \int_0^2 (2 – x)\sqrt{x}\,dx$.

$$I = \int_0^2 \left(2x^{1/2} – x^{3/2}\right)dx = \left[\frac{4}{3}x^{3/2} – \frac{2}{5}x^{5/2}\right]_0^2 = \frac{8\sqrt{2}}{3} – \frac{8\sqrt{2}}{5} = \frac{16\sqrt{2}}{15}$$

10. $\int_0^{\pi/2} \left(2\log\sin x – \log\sin 2x\right)dx$

Answer: Since $\log\sin 2x = \log 2 + \log\sin x + \log\cos x$, the integrand simplifies to $\log\tan x – \log 2$. Now $\int_0^{\pi/2}\log\tan x\,dx = 0$ (by $P_4$, as $\log\tan\!\left(\tfrac{\pi}{2} – x\right) = -\log\tan x$). Therefore

$$I = 0 – \frac{\pi}{2}\log 2 = -\frac{\pi}{2}\log 2$$

11. $\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx$

Answer: $\sin^2 x$ is an even function, so $I = 2\int_0^{\pi/2}\sin^2 x\,dx = 2\cdot\dfrac{\pi}{4} = \dfrac{\pi}{2}$ (using the result of Q1).

12. $\int_0^{\pi} \dfrac{x}{1 + \sin x}\,dx$

Answer: By $P_3$ (with $a = 0$, $b = \pi$), $I = \int_0^{\pi}\dfrac{\pi – x}{1 + \sin x}\,dx$. Adding,

$$2I = \pi\int_0^{\pi}\frac{dx}{1 + \sin x} = \pi\int_0^{\pi}\left(\sec^2 x – \sec x\tan x\right)dx = \pi\left[\frac{\sin x – 1}{\cos x}\right]_0^{\pi} = \pi\big(1 – (-1)\big) = 2\pi$$

Hence $I = \pi$. (The antiderivative $\dfrac{\sin x – 1}{\cos x}$ is continuous through $x = \tfrac{\pi}{2}$.)

13. $\int_{-\pi/2}^{\pi/2} \sin^7 x\,dx$

Answer: $\sin^7(-x) = -\sin^7 x$, so $\sin^7 x$ is an odd function and the integral over the symmetric interval $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ is $0$.

14. $\int_0^{2\pi} \cos^5 x\,dx$

Answer: Using $\int_0^{2a}f(x)\,dx = 2\int_0^a f(x)\,dx$ when $f(2a – x) = f(x)$: here $\cos^5(2\pi – x) = \cos^5 x$, so $I = 2\int_0^{\pi}\cos^5 x\,dx$. But $\cos^5(\pi – x) = -\cos^5 x$, which makes $\int_0^{\pi}\cos^5 x\,dx = 0$. Hence $I = 0$.

15. $\int_0^{\pi/2} \dfrac{\sin x – \cos x}{1 + \sin x\cos x}\,dx$

Answer: By $P_4$, replacing $x$ by $\tfrac{\pi}{2} – x$ interchanges $\sin x$ and $\cos x$, giving $I = \int_0^{\pi/2}\dfrac{\cos x – \sin x}{1 + \sin x\cos x}\,dx = -I$. Therefore $2I = 0$, so $I = 0$.

16. $\int_0^{\pi} \log(1 + \cos x)\,dx$

Answer: By $P_4$ (with $a = \pi$), $I = \int_0^{\pi}\log(1 – \cos x)\,dx$. Adding,

$$2I = \int_0^{\pi}\log\!\big[(1 + \cos x)(1 – \cos x)\big]dx = \int_0^{\pi}\log(\sin^2 x)\,dx = 2\int_0^{\pi}\log\sin x\,dx$$

Now $\int_0^{\pi}\log\sin x\,dx = 2\int_0^{\pi/2}\log\sin x\,dx = 2\left(-\tfrac{\pi}{2}\log 2\right) = -\pi\log 2$. Hence $2I = -2\pi\log 2$, so $I = -\pi\log 2$.

17. $\int_0^a \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{a – x}}\,dx$

Answer: By $P_4$, $I = \int_0^a \dfrac{\sqrt{a – x}}{\sqrt{a – x} + \sqrt{x}}\,dx$. Adding gives $2I = \int_0^a dx = a$, so $I = \dfrac{a}{2}$.

18. $\int_0^4 |x – 1|\,dx$

Answer: Split at $x = 1$.

$$I = \int_0^1 (1 – x)\,dx + \int_1^4 (x – 1)\,dx = \frac{1}{2} + \frac{9}{2} = 5$$

19. Show that $\int_0^a f(x)\,g(x)\,dx = 2\int_0^a f(x)\,dx$, if $f$ and $g$ are defined by $f(x) = f(a – x)$ and $g(x) + g(a – x) = 4$.

Answer: Let $I = \int_0^a f(x)g(x)\,dx$. By $P_4$ and $f(a – x) = f(x)$,

$$I = \int_0^a f(a – x)\,g(a – x)\,dx = \int_0^a f(x)\,g(a – x)\,dx$$

Adding the two expressions for $I$ and using $g(x) + g(a – x) = 4$,

$$2I = \int_0^a f(x)\big[g(x) + g(a – x)\big]dx = 4\int_0^a f(x)\,dx \quad\Rightarrow\quad I = 2\int_0^a f(x)\,dx$$

Choose the correct answer in Exercises 20 and 21.

20. The value of $\int_{-\pi/2}^{\pi/2} \left(x^3 + x\cos x + \tan^5 x + 1\right)dx$ is

(A) 0   (B) 2   (C) $\pi$   (D) 1

Answer: (C) $\pi$. Each of $x^3$, $x\cos x$ and $\tan^5 x$ is an odd function, so their integrals over the symmetric interval vanish. Only the constant $1$ survives, giving $\int_{-\pi/2}^{\pi/2} 1\,dx = \pi$.

21. The value of $\int_0^{\pi/2} \log\!\left(\dfrac{4 + 3\sin x}{4 + 3\cos x}\right)dx$ is

(A) 2   (B) $\tfrac{3}{4}$   (C) 0   (D) $-2$

Answer: (C) 0. By $P_4$, replacing $x$ by $\tfrac{\pi}{2} – x$ swaps $\sin x$ and $\cos x$, so $I = \int_0^{\pi/2}\log\!\left(\dfrac{4 + 3\cos x}{4 + 3\sin x}\right)dx = -I$. Hence $2I = 0$, so $I = 0$.

Miscellaneous Exercise on Chapter 7

Integrate the functions in Exercises 1 to 23.

1. $\dfrac{1}{x – x^3}$

Answer: $x – x^3 = x(1 – x)(1 + x)$. Partial fractions give $\dfrac{1}{x(1 – x)(1 + x)} = \dfrac{1}{x} + \dfrac{1/2}{1 – x} – \dfrac{1/2}{1 + x}$.

$$\int\frac{dx}{x – x^3} = \log|x| – \frac{1}{2}\log|1 – x| – \frac{1}{2}\log|1 + x| + C = \frac{1}{2}\log\left|\frac{x^2}{1 – x^2}\right| + C$$

2. $\dfrac{1}{\sqrt{x + a} + \sqrt{x + b}}$

Answer: Rationalise by multiplying above and below by $\sqrt{x + a} – \sqrt{x + b}$; the denominator becomes $(x + a) – (x + b) = a – b$.

$$\int\frac{dx}{\sqrt{x + a} + \sqrt{x + b}} = \frac{1}{a – b}\int\left(\sqrt{x + a} – \sqrt{x + b}\right)dx = \frac{2}{3(a – b)}\left[(x + a)^{3/2} – (x + b)^{3/2}\right] + C$$

3. $\dfrac{1}{x\sqrt{ax – x^2}}$  (Hint: Put $x = \dfrac{a}{t}$)

Answer: With $x = \dfrac{a}{t}$, $dx = -\dfrac{a}{t^2}\,dt$ and $ax – x^2 = \dfrac{a^2(t – 1)}{t^2}$, so $x\sqrt{ax – x^2} = \dfrac{a^2\sqrt{t – 1}}{t^2}$.

$$\int\frac{dx}{x\sqrt{ax – x^2}} = -\frac{1}{a}\int\frac{dt}{\sqrt{t – 1}} = -\frac{2}{a}\sqrt{t – 1} + C = -\frac{2}{a}\sqrt{\frac{a – x}{x}} + C$$

4. $\dfrac{1}{x^2(x^4 + 1)^{3/4}}$

Answer: Take $x^4$ out of the bracket: $(x^4 + 1)^{3/4} = x^3\left(1 + x^{-4}\right)^{3/4}$, so the integrand is $\dfrac{1}{x^5\left(1 + x^{-4}\right)^{3/4}}$. Put $t = 1 + x^{-4}$, giving $x^{-5}dx = -\tfrac{1}{4}\,dt$.

$$\int\frac{dx}{x^2(x^4 + 1)^{3/4}} = -\frac{1}{4}\int t^{-3/4}\,dt = -t^{1/4} + C = -\left(1 + \frac{1}{x^4}\right)^{1/4} + C = -\frac{(x^4 + 1)^{1/4}}{x} + C$$

5. $\dfrac{1}{x^{1/2} + x^{1/3}}$  (Hint: put $x = t^6$)

Answer: With $x = t^6$, $dx = 6t^5\,dt$ and the denominator is $t^3 + t^2 = t^2(t + 1)$. Since $\dfrac{t^3}{t + 1} = t^2 – t + 1 – \dfrac{1}{t + 1}$,

$$\int\frac{dx}{x^{1/2} + x^{1/3}} = 6\int\frac{t^3}{t + 1}\,dt = 2t^3 – 3t^2 + 6t – 6\log|t + 1| + C$$

Restoring $t = x^{1/6}$: $= 2\sqrt{x} – 3x^{1/3} + 6x^{1/6} – 6\log\!\left(1 + x^{1/6}\right) + C$.

6. $\dfrac{5x}{(x + 1)(x^2 + 9)}$

Answer: Writing $\dfrac{5x}{(x + 1)(x^2 + 9)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 9}$ gives $A = -\tfrac{1}{2}$, $B = \tfrac{1}{2}$, $C = \tfrac{9}{2}$.

$$\int\frac{5x\,dx}{(x + 1)(x^2 + 9)} = -\frac{1}{2}\log|x + 1| + \frac{1}{4}\log(x^2 + 9) + \frac{3}{2}\tan^{-1}\frac{x}{3} + C$$

7. $\dfrac{\sin x}{\sin(x – a)}$

Answer: Write $\sin x = \sin\big((x – a) + a\big) = \sin(x – a)\cos a + \cos(x – a)\sin a$, so $\dfrac{\sin x}{\sin(x – a)} = \cos a + \sin a\cot(x – a)$.

$$\int\frac{\sin x}{\sin(x – a)}\,dx = x\cos a + \sin a\,\log|\sin(x – a)| + C$$

8. $\dfrac{e^{5\log x} – e^{4\log x}}{e^{3\log x} – e^{2\log x}}$

Answer: Since $e^{n\log x} = x^n$, the fraction is $\dfrac{x^5 – x^4}{x^3 – x^2} = \dfrac{x^4(x – 1)}{x^2(x – 1)} = x^2$.

$$\int\frac{e^{5\log x} – e^{4\log x}}{e^{3\log x} – e^{2\log x}}\,dx = \int x^2\,dx = \frac{x^3}{3} + C$$

9. $\dfrac{\cos x}{\sqrt{4 – \sin^2 x}}$

Answer: Put $t = \sin x$, so $dt = \cos x\,dx$.

$$\int\frac{\cos x\,dx}{\sqrt{4 – \sin^2 x}} = \int\frac{dt}{\sqrt{2^2 – t^2}} = \sin^{-1}\frac{t}{2} + C = \sin^{-1}\!\left(\frac{\sin x}{2}\right) + C$$

10. $\dfrac{\sin^8 x – \cos^8 x}{1 – 2\sin^2 x\cos^2 x}$

Answer: Numerator $= (\sin^4 x – \cos^4 x)(\sin^4 x + \cos^4 x) = (\sin^2 x – \cos^2 x)(\sin^4 x + \cos^4 x)$. Denominator $= 1 – 2\sin^2 x\cos^2 x = \sin^4 x + \cos^4 x$. So the fraction equals $\sin^2 x – \cos^2 x = -\cos 2x$.

$$\int\frac{\sin^8 x – \cos^8 x}{1 – 2\sin^2 x\cos^2 x}\,dx = -\int\cos 2x\,dx = -\frac{1}{2}\sin 2x + C$$

11. $\dfrac{1}{\cos(x + a)\cos(x + b)}$

Answer: Multiply and divide by $\sin(a – b)$ and use $\sin(a – b) = \sin\big((x + a) – (x + b)\big)$ to get $\dfrac{1}{\sin(a – b)}\big[\tan(x + a) – \tan(x + b)\big]$.

$$\int\frac{dx}{\cos(x + a)\cos(x + b)} = \frac{1}{\sin(a – b)}\log\left|\frac{\cos(x + b)}{\cos(x + a)}\right| + C$$

12. $\dfrac{x^3}{\sqrt{1 – x^8}}$

Answer: Put $t = x^4$, so $dt = 4x^3\,dx$ and $x^8 = t^2$.

$$\int\frac{x^3\,dx}{\sqrt{1 – x^8}} = \frac{1}{4}\int\frac{dt}{\sqrt{1 – t^2}} = \frac{1}{4}\sin^{-1}(x^4) + C$$

13. $\dfrac{e^x}{(1 + e^x)(2 + e^x)}$

Answer: Put $t = e^x$, so $dt = e^x\,dx$. Then $\dfrac{1}{(1 + t)(2 + t)} = \dfrac{1}{1 + t} – \dfrac{1}{2 + t}$.

$$\int\frac{e^x\,dx}{(1 + e^x)(2 + e^x)} = \log\left|\frac{1 + e^x}{2 + e^x}\right| + C$$

14. $\dfrac{1}{(x^2 + 1)(x^2 + 4)}$

Answer: $\dfrac{1}{(x^2 + 1)(x^2 + 4)} = \dfrac{1}{3}\left(\dfrac{1}{x^2 + 1} – \dfrac{1}{x^2 + 4}\right)$.

$$\int\frac{dx}{(x^2 + 1)(x^2 + 4)} = \frac{1}{3}\tan^{-1}x – \frac{1}{6}\tan^{-1}\frac{x}{2} + C$$

15. $\cos^3 x\,e^{\log\sin x}$

Answer: $e^{\log\sin x} = \sin x$, so the integrand is $\cos^3 x\sin x$. Put $t = \cos x$, $dt = -\sin x\,dx$.

$$\int\cos^3 x\,e^{\log\sin x}\,dx = -\int t^3\,dt = -\frac{\cos^4 x}{4} + C$$

16. $e^{3\log x}(x^4 + 1)^{-1}$

Answer: $e^{3\log x} = x^3$, so the integrand is $\dfrac{x^3}{x^4 + 1}$. Put $t = x^4 + 1$, $dt = 4x^3\,dx$.

$$\int\frac{x^3\,dx}{x^4 + 1} = \frac{1}{4}\log(x^4 + 1) + C$$

17. $f^{\prime}(ax + b)\,[f(ax + b)]^n$

Answer: Put $t = f(ax + b)$, so $dt = a\,f^{\prime}(ax + b)\,dx$.

$$\int f^{\prime}(ax + b)\,[f(ax + b)]^n\,dx = \frac{1}{a}\int t^n\,dt = \frac{[f(ax + b)]^{n+1}}{a(n + 1)} + C \quad (n \neq -1)$$

18. $\dfrac{1}{\sqrt{\sin^3 x\,\sin(x + \alpha)}}$

Answer: Write $\sin^3 x\sin(x + \alpha) = \sin^4 x\cdot\dfrac{\sin(x + \alpha)}{\sin x}$ and $\dfrac{\sin(x + \alpha)}{\sin x} = \cos\alpha + \sin\alpha\cot x$. Put $t = \cos\alpha + \sin\alpha\cot x$, so $\dfrac{dx}{\sin^2 x} = -\dfrac{dt}{\sin\alpha}$.

$$\int\frac{dx}{\sqrt{\sin^3 x\,\sin(x + \alpha)}} = -\frac{1}{\sin\alpha}\int\frac{dt}{\sqrt{t}} = -\frac{2}{\sin\alpha}\sqrt{\frac{\sin(x + \alpha)}{\sin x}} + C$$

19. $\sqrt{\dfrac{1 – \sqrt{x}}{1 + \sqrt{x}}}$

Answer: Multiply inside the root by $\dfrac{1 – \sqrt{x}}{1 – \sqrt{x}}$ to get $\dfrac{1 – \sqrt{x}}{\sqrt{1 – x}}$. Then split and use $x = \sin^2\theta$ for the second piece.

$$\int\sqrt{\frac{1 – \sqrt{x}}{1 + \sqrt{x}}}\,dx = \int\frac{dx}{\sqrt{1 – x}} – \int\frac{\sqrt{x}}{\sqrt{1 – x}}\,dx = \sqrt{x – x^2} – \sin^{-1}\sqrt{x} – 2\sqrt{1 – x} + C$$

20. $\dfrac{2 + \sin 2x}{1 + \cos 2x}\,e^x$

Answer: Using $\sin 2x = 2\sin x\cos x$ and $1 + \cos 2x = 2\cos^2 x$, the factor becomes $\sec^2 x + \tan x$. This is $\tan x + \dfrac{d}{dx}(\tan x)$, so it fits $\int e^x[f(x) + f^{\prime}(x)]dx = e^x f(x)$ with $f(x) = \tan x$.

$$\int\frac{2 + \sin 2x}{1 + \cos 2x}\,e^x\,dx = \int e^x\left(\tan x + \sec^2 x\right)dx = e^x\tan x + C$$

21. $\dfrac{x^2 + x + 1}{(x + 1)^2(x + 2)}$

Answer: Partial fractions: $\dfrac{x^2 + x + 1}{(x + 1)^2(x + 2)} = \dfrac{-2}{x + 1} + \dfrac{1}{(x + 1)^2} + \dfrac{3}{x + 2}$.

$$\int\frac{x^2 + x + 1}{(x + 1)^2(x + 2)}\,dx = 3\log|x + 2| – 2\log|x + 1| – \frac{1}{x + 1} + C$$

22. $\tan^{-1}\sqrt{\dfrac{1 – x}{1 + x}}$

Answer: Put $x = \cos\theta$, so $dx = -\sin\theta\,d\theta$. Then $\sqrt{\dfrac{1 – \cos\theta}{1 + \cos\theta}} = \tan\dfrac{\theta}{2}$, so $\tan^{-1}\sqrt{\dfrac{1 – x}{1 + x}} = \dfrac{\theta}{2}$.

$$\int\tan^{-1}\sqrt{\frac{1 – x}{1 + x}}\,dx = -\frac{1}{2}\int\theta\sin\theta\,d\theta = \frac{x}{2}\cos^{-1}x – \frac{1}{2}\sqrt{1 – x^2} + C$$

23. $\dfrac{\sqrt{x^2 + 1}\,\big[\log(x^2 + 1) – 2\log x\big]}{x^4}$

Answer: $\log(x^2 + 1) – 2\log x = \log\!\left(1 + \tfrac{1}{x^2}\right)$ and $\dfrac{\sqrt{x^2 + 1}}{x^4} = \dfrac{1}{x^3}\sqrt{1 + \tfrac{1}{x^2}}$. Put $t = 1 + \dfrac{1}{x^2}$, so $\dfrac{dx}{x^3} = -\dfrac{dt}{2}$, then integrate $t^{1/2}\log t$ by parts.

$$\int\frac{\sqrt{x^2 + 1}\,\big[\log(x^2 + 1) – 2\log x\big]}{x^4}\,dx = -\frac{1}{3}\left(1 + \frac{1}{x^2}\right)^{3/2}\left[\log\!\left(1 + \frac{1}{x^2}\right) – \frac{2}{3}\right] + C$$

Evaluate the definite integrals in Exercises 24 to 31.

24. $\int_{\pi/2}^{\pi} e^x\left(\dfrac{1 – \sin x}{1 – \cos x}\right)dx$

Answer: Using half-angles, $\dfrac{1 – \sin x}{1 – \cos x} = \dfrac{1}{2}\csc^2\dfrac{x}{2} – \cot\dfrac{x}{2}$, which is $f(x) + f^{\prime}(x)$ with $f(x) = -\cot\dfrac{x}{2}$. Hence the antiderivative is $-e^x\cot\dfrac{x}{2}$.

$$I = \left[-e^x\cot\frac{x}{2}\right]_{\pi/2}^{\pi} = 0 – \left(-e^{\pi/2}\cdot 1\right) = e^{\pi/2}$$

25. $\int_0^{\pi/4} \dfrac{\sin x\cos x}{\cos^4 x + \sin^4 x}\,dx$

Answer: Dividing above and below by $\cos^4 x$ gives $\dfrac{\tan x\sec^2 x}{1 + \tan^4 x}$. Put $t = \tan^2 x$, so $\tan x\sec^2 x\,dx = \tfrac{1}{2}\,dt$; limits $0 \to 0$, $\tfrac{\pi}{4} \to 1$.

$$I = \frac{1}{2}\int_0^1\frac{dt}{1 + t^2} = \frac{1}{2}\big[\tan^{-1}t\big]_0^1 = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$$

26. $\int_0^{\pi/2} \dfrac{\cos^2 x}{\cos^2 x + 4\sin^2 x}\,dx$

Answer: Put $t = \tan x$; then $dx = \dfrac{dt}{1 + t^2}$ and the integrand becomes $\dfrac{1}{1 + 4t^2}$, so $I = \int_0^{\infty}\dfrac{dt}{(1 + t^2)(1 + 4t^2)}$. Using $\dfrac{1}{(1 + t^2)(1 + 4t^2)} = \dfrac{-1/3}{1 + t^2} + \dfrac{4/3}{1 + 4t^2}$,

$$I = -\frac{1}{3}\cdot\frac{\pi}{2} + \frac{4}{3}\cdot\frac{\pi}{4} = -\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6}$$

27. $\int_{\pi/6}^{\pi/3} \dfrac{\sin x + \cos x}{\sqrt{\sin 2x}}\,dx$

Answer: Since $\sin 2x = 1 – (\sin x – \cos x)^2$, put $t = \sin x – \cos x$, so $dt = (\cos x + \sin x)\,dx$. Limits: $x = \tfrac{\pi}{6} \to t = \tfrac{1 – \sqrt{3}}{2}$, $x = \tfrac{\pi}{3} \to t = \tfrac{\sqrt{3} – 1}{2}$.

$$I = \int_{(1 – \sqrt{3})/2}^{(\sqrt{3} – 1)/2}\frac{dt}{\sqrt{1 – t^2}} = \big[\sin^{-1}t\big]_{(1 – \sqrt{3})/2}^{(\sqrt{3} – 1)/2} = 2\sin^{-1}\!\left(\frac{\sqrt{3} – 1}{2}\right)$$

28. $\int_0^1 \dfrac{dx}{\sqrt{1 + x} – \sqrt{x}}$

Answer: Rationalise; the denominator becomes $(1 + x) – x = 1$.

$$I = \int_0^1\left(\sqrt{1 + x} + \sqrt{x}\right)dx = \left[\frac{2}{3}(1 + x)^{3/2} + \frac{2}{3}x^{3/2}\right]_0^1 = \frac{2}{3}\left(2\sqrt{2} + 1\right) – \frac{2}{3} = \frac{4\sqrt{2}}{3}$$

29. $\int_0^{\pi/4} \dfrac{\sin x + \cos x}{9 + 16\sin 2x}\,dx$

Answer: With $t = \sin x – \cos x$, $\sin 2x = 1 – t^2$ so $9 + 16\sin 2x = 25 – 16t^2$. Limits: $x = 0 \to t = -1$, $x = \tfrac{\pi}{4} \to t = 0$.

$$I = \int_{-1}^{0}\frac{dt}{25 – 16t^2} = \frac{1}{40}\left[\log\left|\frac{5 + 4t}{5 – 4t}\right|\right]_{-1}^{0} = \frac{1}{40}\log 9 = \frac{1}{20}\log 3$$

30. $\int_0^{\pi/2} \sin 2x\,\tan^{-1}(\sin x)\,dx$

Answer: Put $t = \sin x$; then $\sin 2x\,dx = 2\sin x\cos x\,dx = 2t\,dt$, and the limits become $0 \to 1$. Integrating $t\tan^{-1}t$ by parts,

$$I = 2\int_0^1 t\tan^{-1}t\,dt = 2\left[\frac{t^2}{2}\tan^{-1}t – \frac{1}{2}\left(t – \tan^{-1}t\right)\right]_0^1 = 2\left(\frac{\pi}{4} – \frac{1}{2}\right) = \frac{\pi}{2} – 1$$

31. $\int_1^4 \big[|x – 1| + |x – 2| + |x – 3|\big]\,dx$

Answer: Evaluate each modulus over $[1, 4]$ by splitting at its breakpoint:

$$\int_1^4 |x – 1|\,dx = \frac{9}{2}, \qquad \int_1^4 |x – 2|\,dx = \frac{5}{2}, \qquad \int_1^4 |x – 3|\,dx = \frac{5}{2}$$

Adding, $I = \dfrac{9}{2} + \dfrac{5}{2} + \dfrac{5}{2} = \dfrac{19}{2}$.

Prove the following (Exercises 32 to 37).

32. $\int_1^3 \dfrac{dx}{x^2(x + 1)} = \dfrac{2}{3} + \log\dfrac{2}{3}$

Answer: Partial fractions: $\dfrac{1}{x^2(x + 1)} = -\dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x + 1}$, so the antiderivative is $\log\left|\dfrac{x + 1}{x}\right| – \dfrac{1}{x}$.

$$I = \left[\log\frac{x + 1}{x} – \frac{1}{x}\right]_1^3 = \left(\log\frac{4}{3} – \frac{1}{3}\right) – \left(\log 2 – 1\right) = \frac{2}{3} + \log\frac{2}{3}$$

33. $\int_0^1 x\,e^x\,dx = 1$

Answer: By parts, $\int x e^x\,dx = e^x(x – 1)$.

$$I = \big[e^x(x – 1)\big]_0^1 = 0 – (-1) = 1$$

34. $\int_{-1}^{1} x^{17}\cos^4 x\,dx = 0$

Answer: Let $f(x) = x^{17}\cos^4 x$. Then $f(-x) = (-x)^{17}\cos^4 x = -f(x)$, so $f$ is odd and the integral over the symmetric interval $[-1, 1]$ is $0$.

35. $\int_0^{\pi/2} \sin^3 x\,dx = \dfrac{2}{3}$

Answer: $\sin^3 x = \sin x(1 – \cos^2 x)$. Put $t = \cos x$, $dt = -\sin x\,dx$.

$$I = \int_0^1 (1 – t^2)\,dt = \left[t – \frac{t^3}{3}\right]_0^1 = 1 – \frac{1}{3} = \frac{2}{3}$$

36. $\int_0^{\pi/4} 2\tan^3 x\,dx = 1 – \log 2$

Answer: $\tan^3 x = \tan x\sec^2 x – \tan x$, so $\int\tan^3 x\,dx = \dfrac{\tan^2 x}{2} + \log|\cos x|$.

$$I = 2\left[\frac{\tan^2 x}{2} + \log|\cos x|\right]_0^{\pi/4} = 2\left(\frac{1}{2} – \frac{1}{2}\log 2\right) = 1 – \log 2$$

37. $\int_0^1 \sin^{-1}x\,dx = \dfrac{\pi}{2} – 1$

Answer: By parts (taking $1$ as the second function), $\int\sin^{-1}x\,dx = x\sin^{-1}x + \sqrt{1 – x^2}$.

$$I = \left[x\sin^{-1}x + \sqrt{1 – x^2}\right]_0^1 = \frac{\pi}{2} – 1$$

Choose the correct answers in Exercises 38 to 40.

38. $\int \dfrac{dx}{e^x + e^{-x}}$ is equal to

(A) $\tan^{-1}(e^x) + C$   (B) $\tan^{-1}(e^{-x}) + C$   (C) $\log(e^x – e^{-x}) + C$   (D) $\log(e^x + e^{-x}) + C$

Answer: (A). Multiplying above and below by $e^x$ gives $\dfrac{e^x}{e^{2x} + 1}$. Put $t = e^x$, $dt = e^x\,dx$:

$$\int\frac{dx}{e^x + e^{-x}} = \int\frac{dt}{t^2 + 1} = \tan^{-1}(e^x) + C$$

39. $\int \dfrac{\cos 2x}{(\sin x + \cos x)^2}\,dx$ is equal to

(A) $\dfrac{-1}{\sin x + \cos x} + C$   (B) $\log|\sin x + \cos x| + C$   (C) $\log|\sin x – \cos x| + C$   (D) $\dfrac{1}{(\sin x + \cos x)^2}$

Answer: (B). Since $\cos 2x = (\cos x – \sin x)(\cos x + \sin x)$, the integrand reduces to $\dfrac{\cos x – \sin x}{\sin x + \cos x}$. Put $t = \sin x + \cos x$, $dt = (\cos x – \sin x)\,dx$:

$$\int\frac{\cos 2x\,dx}{(\sin x + \cos x)^2} = \int\frac{dt}{t} = \log|\sin x + \cos x| + C$$

40. If $f(a + b – x) = f(x)$, then $\int_a^b x\,f(x)\,dx$ is equal to

(A) $\dfrac{a + b}{2}\int_a^b f(b – x)\,dx$   (B) $\dfrac{a + b}{2}\int_a^b f(b + x)\,dx$   (C) $\dfrac{b – a}{2}\int_a^b f(x)\,dx$   (D) $\dfrac{a + b}{2}\int_a^b f(x)\,dx$

Answer: (D). Let $I = \int_a^b x f(x)\,dx$. By $P_3$ and $f(a + b – x) = f(x)$, $I = \int_a^b (a + b – x)f(x)\,dx$. Adding the two,

$$2I = (a + b)\int_a^b f(x)\,dx \quad\Rightarrow\quad I = \frac{a + b}{2}\int_a^b f(x)\,dx$$

Additional Questions and Answers

These extra questions on Integrals reinforce the properties of definite integrals, integration by substitution and the fundamental theorem of calculus for ASSEB Class 12 Mathematics.

Multiple Choice Questions

1. The value of $\int_0^{\pi/2} \dfrac{\cos x}{\sin x + \cos x}\,dx$ is

(A) $\dfrac{\pi}{2}$   (B) $\dfrac{\pi}{4}$   (C) $0$   (D) $\pi$

Answer: (B) $\dfrac{\pi}{4}$. By $P_4$ the integral equals $\int_0^{\pi/2}\dfrac{\sin x}{\cos x + \sin x}\,dx$; adding gives $2I = \int_0^{\pi/2}dx = \dfrac{\pi}{2}$.

2. $\int_{-\pi/4}^{\pi/4} \sin^3 x\,dx$ equals

(A) $1$   (B) $\dfrac{1}{2}$   (C) $0$   (D) $2$

Answer: (C) $0$. $\sin^3 x$ is an odd function, so its integral over the symmetric interval is zero.

3. $\int_a^b f(x)\,dx + \int_b^a f(x)\,dx$ equals

(A) $2\int_a^b f(x)\,dx$   (B) $0$   (C) $\int_a^b f(x)\,dx$   (D) $f(b) – f(a)$

Answer: (B) $0$. Since $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$ (property $P_1$), the sum is zero.

4. $\int_0^2 [x]\,dx$, where $[x]$ is the greatest-integer function, equals

(A) $0$   (B) $1$   (C) $2$   (D) $3$

Answer: (B) $1$. $\int_0^2 [x]\,dx = \int_0^1 0\,dx + \int_1^2 1\,dx = 0 + 1 = 1$.

5. $\int_0^{\pi} \sin x\,dx$ equals

(A) $0$   (B) $1$   (C) $2$   (D) $-2$

Answer: (C) $2$. $\int_0^{\pi}\sin x\,dx = \big[-\cos x\big]_0^{\pi} = -(-1) – (-1) = 2$.

6. If $f$ is an even function, then $\int_{-a}^{a} f(x)\,dx$ equals

(A) $0$   (B) $2\int_0^a f(x)\,dx$   (C) $\int_0^a f(x)\,dx$   (D) $2\int_0^a f(-x)\,dx$

Answer: (B) $2\int_0^a f(x)\,dx$. This is property $P_7(i)$ for even functions.

7. $\int_1^2 \dfrac{1}{x}\,dx$ equals

(A) $\log 2$   (B) $\log 3$   (C) $1$   (D) $\dfrac{1}{2}$

Answer: (A) $\log 2$. $\int_1^2\dfrac{dx}{x} = \big[\log x\big]_1^2 = \log 2 – \log 1 = \log 2$.

8. $\int_{-1}^{1} |x|\,dx$ equals

(A) $0$   (B) $\dfrac{1}{2}$   (C) $1$   (D) $2$

Answer: (C) $1$. $|x|$ is even, so $\int_{-1}^{1}|x|\,dx = 2\int_0^1 x\,dx = 2\cdot\dfrac{1}{2} = 1$.

Fill in the Blanks

1. $\int_0^a f(x)\,dx = \int_0^a f(\underline{\quad})\,dx$.
2. $\int_{-a}^{a} f(x)\,dx = 0$ if $f$ is an $\underline{\quad}$ function.
3. $\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$ provided $f(2a – x) = \underline{\quad}$.
4. By the first fundamental theorem of calculus, $\dfrac{d}{dx}\int_a^x f(t)\,dt = \underline{\quad}$.
5. $\int_a^b f(x)\,dx = -\int_{\underline{\quad}}^{\underline{\quad}} f(x)\,dx$.

Answer: (1) $f(a – x)$;  (2) odd;  (3) $f(x)$;  (4) $f(x)$;  (5) $\int_b^a f(x)\,dx$ (limits $b$ and $a$).

True or False

1. $\int_0^{\pi/2} \sin^2 x\,dx = \int_0^{\pi/2} \cos^2 x\,dx$.
2. $\int_{-2}^{2} x^3\,dx \neq 0$.
3. The value of a definite integral depends on the letter used for the variable of integration.
4. For an odd function $f$, $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$.
5. $\int_0^1 x(1 – x)^{10}\,dx = \dfrac{1}{132}$.

Answer: (1) True (by $P_4$);  (2) False, the integral is $0$ since $x^3$ is odd;  (3) False (property $P_0$);  (4) False, it equals $0$;  (5) True, since $\int_0^1 x(1 – x)^n\,dx = \dfrac{1}{(n + 1)(n + 2)} = \dfrac{1}{132}$ for $n = 10$.

Short Answer Questions

1. Evaluate $\int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\,dx$.

Answer: By $P_4$, $I = \int_0^{\pi/2}\dfrac{\cos x}{\cos x + \sin x}\,dx$. Adding, $2I = \int_0^{\pi/2}dx = \dfrac{\pi}{2}$, so $I = \dfrac{\pi}{4}$.

2. Evaluate $\int_{-1}^{1} \left(x^3 + x^5\right)\cos^2 x\,dx$.

Answer: The integrand is a product of the odd function $x^3 + x^5$ and the even function $\cos^2 x$, hence odd. By property $P_7(ii)$ the integral over $[-1, 1]$ is $0$.

3. Evaluate $\int_0^{\pi} x\sin x\,dx$.

Answer: By $P_3$, $I = \int_0^{\pi}(\pi – x)\sin x\,dx$ (using $\sin(\pi – x) = \sin x$). Adding, $2I = \pi\int_0^{\pi}\sin x\,dx = \pi\cdot 2 = 2\pi$, so $I = \pi$.

4. Show that $\int_0^{\pi/2} \log\tan x\,dx = 0$.

Answer: By $P_4$, replacing $x$ by $\dfrac{\pi}{2} – x$ gives $I = \int_0^{\pi/2}\log\cot x\,dx = \int_0^{\pi/2}\big(-\log\tan x\big)\,dx = -I$. Therefore $2I = 0$, i.e. $I = 0$.

Key Terms

Term	Meaning
Integral (অনুকল)	A function whose derivative is the given function; the inverse process of differentiation.
Indefinite integral	$\int f(x)\,dx = F(x) + C$, containing an arbitrary constant of integration $C$.
Definite integral	$\int_a^b f(x)\,dx$, a number equal to the net signed area under $y = f(x)$ between $x = a$ and $x = b$.
Limits of integration	The lower limit $a$ and upper limit $b$ of a definite integral.
Integration by substitution	Changing the variable of integration to reduce an integral to a standard form.
Integration by parts	$\int u\,v\,dx = u\int v\,dx – \int\left(\frac{du}{dx}\int v\,dx\right)dx$, used for products of functions.
Partial fractions	Splitting a rational function into simpler fractions that can be integrated term by term.
Even function	A function with $f(-x) = f(x)$; then $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$.
Odd function	A function with $f(-x) = -f(x)$; then $\int_{-a}^{a} f(x)\,dx = 0$.
Fundamental theorem of calculus	Links differentiation and integration: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$ and $\int_a^b f(x)\,dx = F(b) – F(a)$.
Properties of definite integrals	Rules $P_0$ to $P_7$ (e.g. $\int_0^a f(x)\,dx = \int_0^a f(a – x)\,dx$) that simplify evaluation.