Complete solutions for Class 12 Maths (NCERT) Exercise 7.2 on Integrals. Master the method of substitution (u-Substitution) to solve complex integrals involving algebraic, trigonometric, and exponential functions, including types like $\int f(g(x))g'(x) dx$ and special trigonometric forms like $\int \frac{1}{1 \pm \tan x} dx$.
This exercise focuses entirely on integration by the method of substitution (or u-substitution), which is essential for transforming complex integrals into standard forms.
1. $\int \frac{2x}{1 + x^2} dx$
Let $u = 1 + x^2$.
Then $du = 2x \, dx$.
$$\int \frac{du}{u} = \log|u| + C = \mathbf{\log|1 + x^2| + C}$$
2. $\int \frac{(\log x)^2}{x} dx$
Let $u = \log x$.
Then $du = \frac{1}{x} \, dx$.
$$\int u^2 du = \frac{u^3}{3} + C = \mathbf{\frac{(\log x)^3}{3} + C}$$
3. $\int \frac{1}{x + x \log x} dx = \int \frac{1}{x(1 + \log x)} dx$
Let $u = 1 + \log x$.
Then $du = \frac{1}{x} \, dx$.
$$\int \frac{du}{u} = \log|u| + C = \mathbf{\log|1 + \log x| + C}$$
4. $\int \sin x \sin(\cos x) dx$
Let $u = \cos x$.
Then $du = -\sin x \, dx$, so $-du = \sin x \, dx$.
$$\int \sin(u) (-du) = -\int \sin u \, du = -(-\cos u) + C = \cos(\cos x) + C$$
$$= \mathbf{\cos(\cos x) + C}$$
5. $\int \sin(ax + b) \cos(ax + b) dx$
Multiply and divide by 2:
$$\int \frac{1}{2} [2 \sin(ax + b) \cos(ax + b)] dx = \frac{1}{2} \int \sin(2(ax + b)) dx$$
Let $u = 2(ax + b)$. Then $du = 2a \, dx$, so $dx = \frac{du}{2a}$.
$$\frac{1}{2} \int \sin(u) \frac{du}{2a} = \frac{1}{4a} (-\cos u) + C$$
$$= \mathbf{-\frac{1}{4a} \cos(2(ax + b)) + C}$$
(Alternatively, use $u = \sin(ax+b)$ or $u = \cos(ax+b)$.)
6. $\int \sqrt{ax + b} dx$
Let $u = ax + b$.
Then $du = a \, dx$, so $dx = \frac{1}{a} du$.
$$\int \sqrt{u} \frac{1}{a} du = \frac{1}{a} \int u^{1/2} du = \frac{1}{a} \frac{u^{3/2}}{3/2} + C$$
$$= \mathbf{\frac{2}{3a} (ax + b)^{3/2} + C}$$
7. $\int x \sqrt{x + 2} dx$
Let $u = x + 2$. Then $du = dx$ and $x = u – 2$.
$$\int (u – 2) \sqrt{u} \, du = \int (u^{3/2} – 2u^{1/2}) du$$
$$= \frac{u^{5/2}}{5/2} – 2 \frac{u^{3/2}}{3/2} + C$$
$$= \mathbf{\frac{2}{5} (x + 2)^{5/2} – \frac{4}{3} (x + 2)^{3/2} + C}$$
8. $\int x \sqrt{1 + 2x^2} dx$
Let $u = 1 + 2x^2$.
Then $du = 4x \, dx$, so $x \, dx = \frac{1}{4} du$.
$$\int \sqrt{u} \frac{1}{4} du = \frac{1}{4} \int u^{1/2} du = \frac{1}{4} \frac{u^{3/2}}{3/2} + C$$
$$= \mathbf{\frac{1}{6} (1 + 2x^2)^{3/2} + C}$$
9. $\int (4x + 2) \sqrt{x^2 + x + 1} dx$
Factor out 2: $\int 2(2x + 1) \sqrt{x^2 + x + 1} dx$.
Let $u = x^2 + x + 1$.
Then $du = (2x + 1) \, dx$.
$$\int 2 \sqrt{u} \, du = 2 \int u^{1/2} du = 2 \frac{u^{3/2}}{3/2} + C$$
$$= \mathbf{\frac{4}{3} (x^2 + x + 1)^{3/2} + C}$$
10. $\int \frac{1}{x – \sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x} – 1)} dx$
Let $u = \sqrt{x} – 1$.
Then $du = \frac{1}{2\sqrt{x}} \, dx$, so $2 \, du = \frac{1}{\sqrt{x}} \, dx$.
$$\int \frac{2 \, du}{u} = 2 \log|u| + C = 2 \log|\sqrt{x} – 1| + C$$
$$= \mathbf{2 \log|\sqrt{x} – 1| + C}$$
11. $\int \frac{x^3}{\sqrt{x^4 + 1}} dx$
Let $u = x^4 + 1$.
Then $du = 4x^3 \, dx$, so $x^3 \, dx = \frac{1}{4} du$.
$$\int \frac{1}{\sqrt{u}} \frac{1}{4} du = \frac{1}{4} \int u^{-1/2} du = \frac{1}{4} \frac{u^{1/2}}{1/2} + C$$
$$= \mathbf{\frac{1}{2} \sqrt{x^4 + 1} + C}$$
12. $\int (x^3 – 1)^{1/3} x^5 dx$
Write $x^5 = x^3 \cdot x^2$.
Let $u = x^3 – 1$. Then $du = 3x^2 \, dx$, so $x^2 \, dx = \frac{1}{3} du$. Also, $x^3 = u + 1$.
$$\int u^{1/3} (u + 1) \frac{1}{3} du = \frac{1}{3} \int (u^{4/3} + u^{1/3}) du$$
$$= \frac{1}{3} \left(\frac{u^{7/3}}{7/3} + \frac{u^{4/3}}{4/3}\right) + C = \frac{1}{3} \left(\frac{3}{7} u^{7/3} + \frac{3}{4} u^{4/3}\right) + C$$
$$= \mathbf{\frac{1}{7} (x^3 – 1)^{7/3} + \frac{1}{4} (x^3 – 1)^{4/3} + C}$$
13. $\int \frac{x^2}{(2 + 3x^3)^3} dx$
Let $u = 2 + 3x^3$.
Then $du = 9x^2 \, dx$, so $x^2 \, dx = \frac{1}{9} du$.
$$\int \frac{1}{u^3} \frac{1}{9} du = \frac{1}{9} \int u^{-3} du = \frac{1}{9} \frac{u^{-2}}{-2} + C$$
$$= -\frac{1}{18 u^2} + C = \mathbf{-\frac{1}{18 (2 + 3x^3)^2} + C}$$
14. $\int \frac{1}{x (\log x)^m} dx$
Let $u = \log x$.
Then $du = \frac{1}{x} \, dx$.
$$\int \frac{1}{u^m} du = \int u^{-m} du = \frac{u^{-m + 1}}{-m + 1} + C$$
$$= \mathbf{\frac{(\log x)^{1 – m}}{1 – m} + C}$$
15. $\int \frac{x}{9 – 4x^2} dx$
Let $u = 9 – 4x^2$.
Then $du = -8x \, dx$, so $x \, dx = -\frac{1}{8} du$.
$$\int \frac{1}{u} \left(-\frac{1}{8}\right) du = -\frac{1}{8} \log|u| + C$$
$$= \mathbf{-\frac{1}{8} \log|9 – 4x^2| + C}$$
16. $\int e^{2x + 3} dx$
Let $u = 2x + 3$.
Then $du = 2 \, dx$, so $dx = \frac{1}{2} du$.
$$\int e^u \frac{1}{2} du = \frac{1}{2} e^u + C = \mathbf{\frac{1}{2} e^{2x + 3} + C}$$
17. $\int \frac{x}{e^{x^2}} dx = \int x e^{-x^2} dx$
Let $u = -x^2$.
Then $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} du$.
$$\int e^u \left(-\frac{1}{2}\right) du = -\frac{1}{2} e^u + C = -\frac{1}{2} e^{-x^2} + C$$
$$= \mathbf{-\frac{1}{2 e^{x^2}} + C}$$
18. $\int \frac{e^{\tan^{-1} x}}{1 + x^2} dx$
Let $u = \tan^{-1} x$.
Then $du = \frac{1}{1 + x^2} \, dx$.
$$\int e^u du = e^u + C = \mathbf{e^{\tan^{-1} x} + C}$$
19. $\int \frac{e^{2x} – 1}{e^{2x} + 1} dx$
Divide numerator and denominator by $e^x$:
$$\int \frac{e^x – e^{-x}}{e^x + e^{-x}} dx$$
Let $u = e^x + e^{-x}$.
Then $du = (e^x – e^{-x}) \, dx$.
$$\int \frac{du}{u} = \log|u| + C = \mathbf{\log|e^x + e^{-x}| + C}$$
20. $\int \frac{e^{2x} – e^{-2x}}{e^{2x} + e^{-2x}} dx$
Let $u = e^{2x} + e^{-2x}$.
Then $du = (2e^{2x} – 2e^{-2x}) \, dx = 2(e^{2x} – e^{-2x}) \, dx$.
So, $(e^{2x} – e^{-2x}) \, dx = \frac{1}{2} du$.
$$\int \frac{1}{u} \frac{1}{2} du = \frac{1}{2} \log|u| + C$$
$$= \mathbf{\frac{1}{2} \log|e^{2x} + e^{-2x}| + C}$$
21. $\int \tan^2(2x – 3) dx$
Use the identity $\tan^2 \theta = \sec^2 \theta – 1$:
$$\int [\sec^2(2x – 3) – 1] dx$$
Let $u = 2x – 3$. Then $du = 2 \, dx$, so $dx = \frac{1}{2} du$.
$$\int (\sec^2 u – 1) \frac{1}{2} du = \frac{1}{2} (\tan u – u) + C$$
$$= \mathbf{\frac{1}{2} \tan(2x – 3) – \frac{1}{2} (2x – 3) + C}$$
22. $\int \sec^2(7 – 4x) dx$
Let $u = 7 – 4x$.
Then $du = -4 \, dx$, so $dx = -\frac{1}{4} du$.
$$\int \sec^2(u) \left(-\frac{1}{4}\right) du = -\frac{1}{4} \tan u + C$$
$$= \mathbf{-\frac{1}{4} \tan(7 – 4x) + C}$$
23. $\int \frac{\sin^{-1} x}{\sqrt{1 – x^2}} dx$
Let $u = \sin^{-1} x$.
Then $du = \frac{1}{\sqrt{1 – x^2}} \, dx$.
$$\int u \, du = \frac{u^2}{2} + C = \mathbf{\frac{1}{2} (\sin^{-1} x)^2 + C}$$
24. $\int \frac{2 \cos x – 3 \sin x}{6 \cos x + 4 \sin x} dx$
Factor the denominator: $\int \frac{2 \cos x – 3 \sin x}{2 (3 \cos x + 2 \sin x)} dx$.
Let $u = 3 \cos x + 2 \sin x$.
Then $du = (-3 \sin x + 2 \cos x) \, dx$.
$$\int \frac{1}{2} \frac{du}{u} = \frac{1}{2} \log|u| + C$$
$$= \mathbf{\frac{1}{2} \log|3 \cos x + 2 \sin x| + C}$$
25. $\int \frac{1}{\cos^2 x (1 – \tan x)^2} dx$
Use $\frac{1}{\cos^2 x} = \sec^2 x$:
$$\int \frac{\sec^2 x}{(1 – \tan x)^2} dx$$
Let $u = 1 – \tan x$.
Then $du = -\sec^2 x \, dx$, so $-du = \sec^2 x \, dx$.
$$\int \frac{-du}{u^2} = -\int u^{-2} du = – \frac{u^{-1}}{-1} + C = \frac{1}{u} + C$$
$$= \mathbf{\frac{1}{1 – \tan x} + C}$$
26. $\int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$
Let $u = \sqrt{x}$.
Then $du = \frac{1}{2\sqrt{x}} \, dx$, so $2 \, du = \frac{1}{\sqrt{x}} \, dx$.
$$\int \cos u (2 \, du) = 2 \sin u + C = \mathbf{2 \sin \sqrt{x} + C}$$
27. $\int \sin(2x) \cos(2x) dx$
Let $u = \sin(2x)$.
Then $du = 2 \cos(2x) \, dx$, so $\cos(2x) \, dx = \frac{1}{2} du$.
$$\int u \frac{1}{2} du = \frac{1}{2} \frac{u^2}{2} + C = \frac{1}{4} \sin^2(2x) + C$$
$$= \mathbf{\frac{1}{4} \sin^2(2x) + C}$$
28. $\int \frac{\cos x}{\sqrt{1 + \sin x}} dx$
Let $u = 1 + \sin x$.
Then $du = \cos x \, dx$.
$$\int \frac{du}{\sqrt{u}} = \int u^{-1/2} du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$$
$$= \mathbf{2\sqrt{1 + \sin x} + C}$$
29. $\int \cot x \log(\sin x) dx$
Let $u = \log(\sin x)$.
Then $du = \frac{1}{\sin x} (\cos x) \, dx = \cot x \, dx$.
$$\int u \, du = \frac{u^2}{2} + C = \mathbf{\frac{1}{2} (\log(\sin x))^2 + C}$$
30. $\int \frac{\sin x}{1 + \cos x} dx$
Let $u = 1 + \cos x$.
Then $du = -\sin x \, dx$, so $-du = \sin x \, dx$.
$$\int \frac{-du}{u} = -\log|u| + C = -\log|1 + \cos x| + C$$
$$= \mathbf{-\log|1 + \cos x| + C}$$
31. $\int \frac{\sin x}{(1 + \cos x)^2} dx$
Let $u = 1 + \cos x$.
Then $du = -\sin x \, dx$, so $-du = \sin x \, dx$.
$$\int \frac{-du}{u^2} = -\int u^{-2} du = – \frac{u^{-1}}{-1} + C = \frac{1}{u} + C$$
$$= \mathbf{\frac{1}{1 + \cos x} + C}$$
32. $\int \frac{1}{1 + \cot x} dx = \int \frac{\sin x}{\sin x + \cos x} dx$
This requires a trick. Multiply numerator by 2 and manipulate:
$$\frac{1}{2} \int \frac{2 \sin x}{\sin x + \cos x} dx = \frac{1}{2} \int \frac{(\sin x + \cos x) + (\sin x – \cos x)}{\sin x + \cos x} dx$$
$$= \frac{1}{2} \int \left(1 + \frac{\sin x – \cos x}{\sin x + \cos x}\right) dx$$
$$= \frac{1}{2} \left[ \int 1 dx – \int \frac{\cos x – \sin x}{\sin x + \cos x} dx \right]$$
In the second integral, let $u = \sin x + \cos x$, so $du = (\cos x – \sin x) \, dx$.
$$\frac{1}{2} \left[ x – \int \frac{du}{u} \right] = \frac{1}{2} [x – \log|u|] + C$$
$$= \mathbf{\frac{1}{2} [x – \log|\sin x + \cos x|] + C}$$
33. $\int \frac{1}{1 – \tan x} dx = \int \frac{\cos x}{\cos x – \sin x} dx$
Similar trick to Q32. Multiply numerator by 2 and manipulate:
$$\frac{1}{2} \int \frac{2 \cos x}{\cos x – \sin x} dx = \frac{1}{2} \int \frac{(\cos x – \sin x) + (\cos x + \sin x)}{\cos x – \sin x} dx$$
$$= \frac{1}{2} \int \left(1 + \frac{\cos x + \sin x}{\cos x – \sin x}\right) dx$$
$$= \frac{1}{2} \left[ \int 1 dx – \int \frac{\sin x + \cos x}{\sin x – \cos x} dx \right]$$
In the second integral, let $u = \cos x – \sin x$, so $du = (-\sin x – \cos x) \, dx = -(\sin x + \cos x) \, dx$.
$$\frac{1}{2} \left[ x – \int \frac{-du}{u} \right] = \frac{1}{2} [x + \log|u|] + C$$
$$= \mathbf{\frac{1}{2} [x + \log|\cos x – \sin x|] + C}$$
34. $\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$
Divide numerator and denominator by $\cos^2 x$:
$$\int \frac{\sqrt{\tan x} / \cos^2 x}{(\sin x \cos x) / \cos^2 x} dx = \int \frac{\sqrt{\tan x} \sec^2 x}{\tan x} dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$$
Let $u = \tan x$.
Then $du = \sec^2 x \, dx$.
$$\int \frac{du}{\sqrt{u}} = \int u^{-1/2} du = 2 u^{1/2} + C = 2 \sqrt{\tan x} + C$$
$$= \mathbf{2 \sqrt{\tan x} + C}$$
35. $\int \frac{(1 + \log x)^2}{x} dx$
Let $u = 1 + \log x$.
Then $du = \frac{1}{x} \, dx$.
$$\int u^2 du = \frac{u^3}{3} + C = \mathbf{\frac{(1 + \log x)^3}{3} + C}$$
36. $\int \frac{(x + 1)(x + \log x)^2}{x} dx$
Rewrite the integrand:
$$\int \left(\frac{x + 1}{x}\right) (x + \log x)^2 dx = \int \left(1 + \frac{1}{x}\right) (x + \log x)^2 dx$$
Let $u = x + \log x$.
Then $du = \left(1 + \frac{1}{x}\right) \, dx$.
$$\int u^2 du = \frac{u^3}{3} + C = \mathbf{\frac{1}{3} (x + \log x)^3 + C}$$
37. $\int \frac{x^3 \sin(\tan^{-1} x^4)}{1 + x^8} dx$
Let $u = \tan^{-1} x^4$.
Then $du = \frac{1}{1 + (x^4)^2} \cdot 4x^3 \, dx = \frac{4x^3}{1 + x^8} \, dx$.
So, $\frac{x^3}{1 + x^8} \, dx = \frac{1}{4} du$.
$$\int \sin u \left(\frac{1}{4} du\right) = \frac{1}{4} (-\cos u) + C$$
$$= \mathbf{-\frac{1}{4} \cos(\tan^{-1} x^4) + C}$$
Multiple Choice Questions
38. $\int \frac{10x^9 + 10^x \log 10}{x^{10} + 10^x} dx$
Let $u = x^{10} + 10^x$.
The derivative of the denominator is:
$$du = \left(\frac{d}{dx} x^{10} + \frac{d}{dx} 10^x\right) dx = (10x^9 + 10^x \log 10) dx$$
The numerator is exactly $du$.
$$\int \frac{du}{u} = \log|u| + C = \log|x^{10} + 10^x| + C$$
The correct answer is (D) $\log (10^x + x^{10}) + C$.
39. $\int \frac{dx}{\sin^2 x \cos^2 x}$
Use the identity $1 = \sin^2 x + \cos^2 x$ in the numerator:
$$\int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx = \int \left(\frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x}\right) dx$$
$$= \int \left(\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}\right) dx = \int (\sec^2 x + \csc^2 x) dx$$
$$= \tan x – \cot x + C$$
The correct answer is (B) $\tan x – \cot x + C$.