NCERT Solutions Class 12 Maths Chapter 7 Ex 7.2 | LearnCBSEHub

Integration by Substitution

NCERT EXERCISE 7.2 • FULL SOLUTIONS Q1-Q39

💡 The Method of Substitution

To integrate complex functions, we change the variable of integration from $x$ to $t$ using a substitution $x = g(t)$ or $g(x) = t$.

Key Step: If we substitute $g(x) = t$, then we must also replace $dx$ by finding $g'(x) dx = dt$.

$$ I = \int f(g(x)) g'(x) dx = \int f(t) dt $$
Questions 01 — 05
Basic substitutions.
1. $\int \frac{2x}{1+x^2} dx$
Let $1+x^2 = t \implies 2x dx = dt$.
$\int \frac{dt}{t} = \log|t| + C = \log|1+x^2| + C$.
2. $\int \frac{(\log x)^2}{x} dx$
Let $\log x = t \implies \frac{1}{x} dx = dt$.
$\int t^2 dt = \frac{t^3}{3} + C = \frac{(\log x)^3}{3} + C$.
3. $\int \frac{1}{x + x\log x} dx$
$\int \frac{1}{x(1+\log x)} dx$. Let $1+\log x = t \implies \frac{1}{x} dx = dt$.
$\int \frac{dt}{t} = \log|t| + C = \log|1+\log x| + C$.
4. $\int \sin x \sin(\cos x) dx$
Let $\cos x = t \implies -\sin x dx = dt \implies \sin x dx = -dt$.
$-\int \sin t dt = -(-\cos t) + C = \cos(\cos x) + C$.
5. $\int \sin(ax+b)\cos(ax+b) dx$
Multiply/divide by 2: $\frac{1}{2} \int \sin(2(ax+b)) dx$.
Let $2(ax+b) = t \implies 2a dx = dt \implies dx = dt/2a$.
$= \frac{1}{2} \int \sin t \frac{dt}{2a} = \frac{1}{4a} (-\cos t) = -\frac{1}{4a}\cos(2ax+2b) + C$.
Questions 06 — 10
Roots and algebraic functions.
6. $\int \sqrt{ax+b} dx$
$= \frac{(ax+b)^{3/2}}{a \cdot \frac{3}{2}} + C = \frac{2}{3a}(ax+b)^{3/2} + C$.
7. $\int x\sqrt{x+2} dx$
Let $x+2 = t \implies dx = dt, x = t-2$.
$\int (t-2)\sqrt{t} dt = \int (t^{3/2} – 2t^{1/2}) dt = \frac{2}{5}t^{5/2} – \frac{4}{3}t^{3/2} + C$.
$= \frac{2}{5}(x+2)^{5/2} – \frac{4}{3}(x+2)^{3/2} + C$.
8. $\int x\sqrt{1+2x^2} dx$
Let $1+2x^2 = t \implies 4x dx = dt \implies x dx = \frac{dt}{4}$.
$\frac{1}{4} \int \sqrt{t} dt = \frac{1}{4} \cdot \frac{2}{3} t^{3/2} = \frac{1}{6}(1+2x^2)^{3/2} + C$.
9. $\int (4x+2)\sqrt{x^2+x+1} dx$
$2 \int (2x+1)\sqrt{x^2+x+1} dx$. Let $x^2+x+1 = t \implies (2x+1)dx = dt$.
$2 \int \sqrt{t} dt = \frac{4}{3}t^{3/2} = \frac{4}{3}(x^2+x+1)^{3/2} + C$.
10. $\int \frac{1}{x – \sqrt{x}} dx$
$\int \frac{1}{\sqrt{x}(\sqrt{x}-1)} dx$. Let $\sqrt{x}-1 = t \implies \frac{1}{2\sqrt{x}} dx = dt$.
$2 \int \frac{dt}{t} = 2\log|t| + C = 2\log|\sqrt{x}-1| + C$.
Questions 11 — 17
Algebraic and Exponential functions.
11. $\int \frac{x}{\sqrt{x+4}} dx$
Let $x+4=t$. $\int \frac{t-4}{\sqrt{t}} dt = \int (t^{1/2} – 4t^{-1/2}) dt$.
$= \frac{2}{3}(x+4)^{3/2} – 8\sqrt{x+4} + C$.
12. $\int (x^3-1)^{1/3}x^5 dx$
Let $x^3-1=t \implies x^3=t+1, 3x^2 dx = dt$.
$\int t^{1/3}(t+1) \frac{dt}{3} = \frac{1}{3} \int (t^{4/3} + t^{1/3}) dt$.
$= \frac{1}{7}(x^3-1)^{7/3} + \frac{1}{4}(x^3-1)^{4/3} + C$.
13. $\int \frac{x^2}{(2+3x^3)^3} dx$
Let $2+3x^3=t \implies 9x^2 dx = dt$.
$\frac{1}{9} \int t^{-3} dt = \frac{1}{9} (\frac{t^{-2}}{-2}) = -\frac{1}{18(2+3x^3)^2} + C$.
14. $\int \frac{1}{x(\log x)^m} dx$
Let $\log x = t \implies \frac{1}{x} dx = dt$.
$\int t^{-m} dt = \frac{t^{1-m}}{1-m} + C = \frac{(\log x)^{1-m}}{1-m} + C$.
15. $\int \frac{x}{9-4x^2} dx$
Let $9-4x^2 = t \implies -8x dx = dt \implies x dx = -dt/8$.
$-\frac{1}{8} \int \frac{dt}{t} = -\frac{1}{8}\log|9-4x^2| + C$.
16. $\int e^{2x+3} dx$
$= \frac{1}{2}e^{2x+3} + C$.
17. $\int \frac{x}{e^{x^2}} dx$
Let $x^2 = t \implies 2x dx = dt$.
$\frac{1}{2} \int e^{-t} dt = -\frac{1}{2}e^{-t} = -\frac{1}{2e^{x^2}} + C$.
Questions 18 — 20
Trigonometric and Exponential manipulations.
18. $\int \frac{e^{\tan^{-1}x}}{1+x^2} dx$
Let $\tan^{-1}x = t \implies \frac{1}{1+x^2} dx = dt$.
$\int e^t dt = e^t + C = e^{\tan^{-1}x} + C$.
19. $\int \frac{e^{2x}-1}{e^{2x}+1} dx$
Divide num and den by $e^x$: $\int \frac{e^x – e^{-x}}{e^x + e^{-x}} dx$.
Let $e^x + e^{-x} = t \implies (e^x – e^{-x}) dx = dt$.
$\int \frac{dt}{t} = \log|e^x + e^{-x}| + C$.
20. $\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} dx$
Let $e^{2x}+e^{-2x} = t \implies (2e^{2x} – 2e^{-2x}) dx = dt \implies (e^{2x}-e^{-2x})dx = dt/2$.
$\frac{1}{2} \int \frac{dt}{t} = \frac{1}{2}\log|e^{2x}+e^{-2x}| + C$.
Questions 21 — 30
Trigonometric Substitutions.
21. $\int \tan^2(2x-3) dx$
Use $\tan^2 \theta = \sec^2 \theta – 1$.
$\int (\sec^2(2x-3) – 1) dx = \frac{1}{2}\tan(2x-3) – x + C$.
22. $\int \sec^2(7-4x) dx$
$= \frac{\tan(7-4x)}{-4} + C = -\frac{1}{4}\tan(7-4x) + C$.
23. $\int \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx$
Let $\sin^{-1}x = t \implies \frac{1}{\sqrt{1-x^2}} dx = dt$.
$\int t dt = \frac{t^2}{2} + C = \frac{1}{2}(\sin^{-1}x)^2 + C$.
24. $\int \frac{2\cos x – 3\sin x}{6\cos x + 4\sin x} dx$
Denominator = $2(3\cos x + 2\sin x)$. Let $3\cos x + 2\sin x = t \implies (-3\sin x + 2\cos x)dx = dt$.
$\frac{1}{2} \int \frac{dt}{t} = \frac{1}{2}\log|3\cos x + 2\sin x| + C$.
25. $\int \frac{1}{\cos^2x(1-\tan x)^2} dx$
$= \int \frac{\sec^2 x}{(1-\tan x)^2} dx$. Let $1-\tan x = t \implies -\sec^2 x dx = dt$.
$-\int t^{-2} dt = -(\frac{-1}{t}) = \frac{1}{1-\tan x} + C$.
26. $\int \frac{\cos\sqrt{x}}{\sqrt{x}} dx$
Let $\sqrt{x} = t \implies \frac{1}{2\sqrt{x}} dx = dt$.
$2 \int \cos t dt = 2\sin t = 2\sin\sqrt{x} + C$.
27. $\int \sqrt{\sin 2x} \cos 2x dx$
Let $\sin 2x = t \implies 2\cos 2x dx = dt \implies \cos 2x dx = dt/2$.
$\frac{1}{2} \int t^{1/2} dt = \frac{1}{2} \frac{t^{3/2}}{3/2} = \frac{1}{3}(\sin 2x)^{3/2} + C$.
28. $\int \frac{\cos x}{\sqrt{1+\sin x}} dx$
Let $1+\sin x = t \implies \cos x dx = dt$.
$\int t^{-1/2} dt = 2t^{1/2} = 2\sqrt{1+\sin x} + C$.
29. $\int \cot x \log(\sin x) dx$
Let $\log(\sin x) = t \implies \frac{1}{\sin x}\cos x dx = dt \implies \cot x dx = dt$.
$\int t dt = \frac{1}{2}(\log(\sin x))^2 + C$.
30. $\int \frac{\sin x}{1+\cos x} dx$
Let $1+\cos x = t \implies -\sin x dx = dt$.
$-\int \frac{dt}{t} = -\log|1+\cos x| + C$.
Questions 31 — 37
Complex Manipulations.
31. $\int \frac{\sin x}{(1+\cos x)^2} dx$
Let $1+\cos x = t \implies -\sin x dx = dt$.
$-\int t^{-2} dt = -(\frac{-1}{t}) = \frac{1}{1+\cos x} + C$.
32. $\int \frac{1}{1+\cot x} dx$
$= \int \frac{\sin x}{\sin x + \cos x} dx$. Divide by 2: $\frac{1}{2} \int \frac{2\sin x}{\sin x + \cos x} dx$.
$= \frac{1}{2} \int \frac{(\sin x + \cos x) + (\sin x – \cos x)}{\sin x + \cos x} dx$
$= \frac{1}{2} \int \left(1 – \frac{\cos x – \sin x}{\sin x + \cos x}\right) dx$
$= \frac{1}{2}x – \frac{1}{2}\log|\sin x + \cos x| + C$.
33. $\int \frac{1}{1-\tan x} dx$
Similar to Q32: $= \int \frac{\cos x}{\cos x – \sin x} dx$.
$= \frac{1}{2} \int \frac{(\cos x – \sin x) + (\cos x + \sin x)}{\cos x – \sin x} dx$
$= \frac{1}{2}x – \frac{1}{2}\log|\cos x – \sin x| + C$.
34. $\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$
Divide numerator and denominator by $\cos^2 x$.
$= \int \frac{\sqrt{\tan x} \sec^2 x}{\tan x} dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$.
Let $\tan x = t \implies \sec^2 x dx = dt$.
$\int t^{-1/2} dt = 2\sqrt{t} = 2\sqrt{\tan x} + C$.
35. $\int \frac{(1+\log x)^2}{x} dx$
Let $1+\log x = t \implies \frac{1}{x} dx = dt$.
$\int t^2 dt = \frac{1}{3}(1+\log x)^3 + C$.
36. $\int \frac{(x+1)(x+\log x)^2}{x} dx$
$= \int (1 + \frac{1}{x})(x+\log x)^2 dx$. Let $x+\log x = t \implies (1+\frac{1}{x}) dx = dt$.
$\int t^2 dt = \frac{1}{3}(x+\log x)^3 + C$.
37. $\int \frac{x^3 \sin(\tan^{-1}x^4)}{1+x^8} dx$
Let $\tan^{-1}x^4 = t \implies \frac{1}{1+(x^4)^2} \cdot 4x^3 dx = dt \implies \frac{x^3}{1+x^8} dx = \frac{dt}{4}$.
$\frac{1}{4} \int \sin t dt = -\frac{1}{4}\cos t = -\frac{1}{4}\cos(\tan^{-1}x^4) + C$.
Questions 38 — 39
Multiple Choice Questions.
38. $\int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x} dx$
Let $x^{10} + 10^x = t$.
Derivative is $(10x^9 + 10^x \log_e 10) dx = dt$.
$\int \frac{dt}{t} = \log|t| + C = \log|x^{10} + 10^x| + C$.
Correct Option: (D)
39. $\int \frac{dx}{\sin^2 x \cos^2 x}$
Replace $1$ in numerator with $\sin^2 x + \cos^2 x$.
$= \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx = \int (\sec^2 x + \text{cosec}^2 x) dx$
$= \tan x – \cot x + C$.
Correct Option: (B)
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