Exercise 7.6 Class 12 Maths Integrals

NCERT Solutions Class 12 Maths Chapter 7 Ex 7.6 | LearnCBSEHub

💡 The Product Rule for Integrals

Integration by Parts is used to integrate the product of two functions.

$$ \int u v \, dx = u \int v \, dx – \int \left( \frac{du}{dx} \int v \, dx \right) dx $$

ILATE Rule: Order of choosing $u$ (first function):
Inverse Trig $\to$ Logarithmic $\to$ Algebraic $\to$ Trigonometric $\to$ Exponential.

Special Property: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$

Questions 01 — 05

Basic Application of ILATE.

1. $\int x \sin x dx$

Algebraic ($x$) is $u$, Trig ($\sin x$) is $v$.
$= x(-\cos x) – \int 1 \cdot (-\cos x) dx = -x\cos x + \int \cos x dx$.
$= -x\cos x + \sin x + C$.

2. $\int x \sin 3x dx$

$u=x, v=\sin 3x$.
$= x(\frac{-\cos 3x}{3}) – \int 1 \cdot (\frac{-\cos 3x}{3}) dx$.
$= -\frac{x}{3}\cos 3x + \frac{1}{3} \int \cos 3x dx$.
Result: $-\frac{x}{3}\cos 3x + \frac{1}{9}\sin 3x + C$.

3. $\int x^2 e^x dx$

$u=x^2, v=e^x$.
$= x^2 e^x – \int 2x e^x dx$. Apply parts again on $\int x e^x dx$.
$= x^2 e^x – 2(x e^x – e^x) + C = e^x(x^2 – 2x + 2) + C$.

4. $\int x \log x dx$

ILATE: Logarithmic is $u$, Algebraic is $v$. $u=\log x, v=x$.
$= \log x \cdot \frac{x^2}{2} – \int \frac{1}{x} \cdot \frac{x^2}{2} dx$.
$= \frac{x^2}{2}\log x – \frac{1}{2} \int x dx = \frac{x^2}{2}\log x – \frac{x^2}{4} + C$.

5. $\int x \log 2x dx$

$u=\log 2x, v=x$.
$= (\log 2x)\frac{x^2}{2} – \int \frac{1}{2x} \cdot 2 \cdot \frac{x^2}{2} dx = \frac{x^2}{2}\log 2x – \int \frac{x}{2} dx$.
$= \frac{x^2}{2}\log 2x – \frac{x^2}{4} + C$.

Questions 06 — 10

Logarithmic and Inverse Trigonometric Functions.

6. $\int x^2 \log x dx$

$u=\log x, v=x^2$.
$= (\log x)\frac{x^3}{3} – \int \frac{1}{x} \cdot \frac{x^3}{3} dx = \frac{x^3}{3}\log x – \frac{1}{3}\int x^2 dx$.
Result: $\frac{x^3}{3}\log x – \frac{x^3}{9} + C$.

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

$u=\sin^{-1}x, v=x$.
$= \frac{x^2}{2}\sin^{-1}x – \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} dx$.
For integral: Let $x^2 = 1 – (1-x^2)$. Split into $\frac{1}{\sqrt{1-x^2}} – \sqrt{1-x^2}$.
$= \frac{x^2}{2}\sin^{-1}x – \frac{1}{2} [ -\sin^{-1}x + \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}x ]$.
Ans: $\frac{(2x^2-1)}{4}\sin^{-1}x + \frac{x}{4}\sqrt{1-x^2} + C$.

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

$u=\tan^{-1}x, v=x$.
$= \frac{x^2}{2}\tan^{-1}x – \frac{1}{2} \int \frac{x^2}{1+x^2} dx$.
Write $x^2 = (x^2+1) – 1$. Integral becomes $1 – \frac{1}{1+x^2}$.
Ans: $\frac{x^2}{2}\tan^{-1}x – \frac{1}{2}(x – \tan^{-1}x) + C$.

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

Similar to Q7. $u=\cos^{-1}x$.
Ans: $\frac{(2x^2-1)}{4}\cos^{-1}x – \frac{x}{4}\sqrt{1-x^2} + C$.

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

Let $\sin^{-1}x = \theta \implies x = \sin\theta, dx = \cos\theta d\theta$.
$\int \theta^2 \cos\theta d\theta$. Apply parts twice (Algebraic $\theta^2$, Trig $\cos\theta$).
$= \theta^2\sin\theta – 2\int \theta\sin\theta d\theta = \theta^2\sin\theta – 2(-\theta\cos\theta + \sin\theta)$.
Replace $\theta$: $x(\sin^{-1}x)^2 + 2\sqrt{1-x^2}\sin^{-1}x – 2x + C$.

Questions 11 — 15

Trig and Log variations.

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

Let $\cos^{-1}x = t \implies \frac{-1}{\sqrt{1-x^2}}dx = dt$. $x=\cos t$.
$-\int t \cos t dt = -(t\sin t – \int \sin t dt) = -t\sin t – \cos t$.
Ans: $-\sqrt{1-x^2}\cos^{-1}x – x + C$.

12. $\int x \sec^2 x dx$

$u=x, v=\sec^2 x$.
$= x \tan x – \int \tan x dx = x \tan x – \log|\sec x| + C$.

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

Treat as product: $\int 1 \cdot \tan^{-1}x dx$. $u=\tan^{-1}x, v=1$.
$= x \tan^{-1}x – \int \frac{x}{1+x^2} dx$.
Ans: $x \tan^{-1}x – \frac{1}{2}\log(1+x^2) + C$.

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

$u=(\log x)^2, v=x$.
$= \frac{x^2}{2}(\log x)^2 – \int 2\log x \cdot \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2}{2}(\log x)^2 – \int x \log x dx$.
Ans: $\frac{x^2}{2}(\log x)^2 – \frac{x^2}{2}\log x + \frac{x^2}{4} + C$.

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

$u=\log x, v=x^2+1$.
$= (\frac{x^3}{3}+x)\log x – \int (\frac{x^3}{3}+x)\frac{1}{x} dx$.
$= (\frac{x^3}{3}+x)\log x – \int (\frac{x^2}{3}+1) dx$.
Ans: $(\frac{x^3}{3}+x)\log x – \frac{x^3}{9} – x + C$.

Questions 16 — 22

Integrals of type $\int e^x [f(x) + f'(x)] dx$.

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

Here $f(x) = \sin x, f'(x) = \cos x$.
Ans: $e^x \sin x + C$.

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

Write $x = (x+1) – 1$.
$\int e^x [\frac{x+1}{(1+x)^2} – \frac{1}{(1+x)^2}] dx = \int e^x [\frac{1}{1+x} + \frac{-1}{(1+x)^2}] dx$.
$f(x) = \frac{1}{1+x}$. Ans: $\frac{e^x}{1+x} + C$.

18. $\int e^x \frac{1+\sin x}{1+\cos x} dx$

Use half angles: $\frac{1+2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{1}{2}\sec^2(x/2) + \tan(x/2)$.
$\int e^x [\tan(x/2) + \frac{1}{2}\sec^2(x/2)] dx$. $f(x) = \tan(x/2)$.
Ans: $e^x \tan(x/2) + C$.

19. $\int e^x (\frac{1}{x} – \frac{1}{x^2}) dx$

$f(x) = 1/x, f'(x) = -1/x^2$.
Ans: $\frac{e^x}{x} + C$.

20. $\int \frac{(x-3)e^x}{(x-1)^3} dx$

Write $x-3 = (x-1) – 2$.
$\int e^x [\frac{x-1}{(x-1)^3} – \frac{2}{(x-1)^3}] dx = \int e^x [\frac{1}{(x-1)^2} + \frac{-2}{(x-1)^3}] dx$.
$f(x) = (x-1)^{-2}$. Ans: $\frac{e^x}{(x-1)^2} + C$.

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

Let $I = \int e^{2x} \sin x dx$. Apply parts: $\sin x \frac{e^{2x}}{2} – \int \cos x \frac{e^{2x}}{2} dx$.
$I = \frac{e^{2x}}{2}\sin x – \frac{1}{2} [ \cos x \frac{e^{2x}}{2} – \int (-\sin x) \frac{e^{2x}}{2} dx ]$.
$I = \frac{e^{2x}}{2}\sin x – \frac{e^{2x}}{4}\cos x – \frac{1}{4}I$.
$\frac{5}{4}I = \frac{e^{2x}}{4}(2\sin x – \cos x)$. Ans: $\frac{e^{2x}}{5}(2\sin x – \cos x) + C$.

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

Put $x=\tan\theta \implies dx=\sec^2\theta d\theta$.
$\int \sin^{-1}(\sin 2\theta) \sec^2\theta d\theta = \int 2\theta \sec^2\theta d\theta$.
Parts: $2[\theta \tan\theta – \log|\sec\theta|]$.
Ans: $2x \tan^{-1}x – \log(1+x^2) + C$.

Questions 23 — 24

Multiple Choice Questions.

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

Put $x^3 = t \implies 3x^2 dx = dt$.
$\frac{1}{3} \int e^t dt = \frac{1}{3}e^t = \frac{1}{3}e^{x^3}$.

Correct Option: (A)

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

$= \int e^x (\sec x + \sec x \tan x) dx$.
$f(x) = \sec x, f'(x) = \sec x \tan x$.
Result: $e^x \sec x + C$.

Correct Option: (B)