RD Sharma Class 12 Exercise 18.26 Indefinite integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 18.26 Indefinite integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.26 Indefinite integrals Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 24 Jan 2022, 09:43 AM IST

The RD Sharma solutions are very well known for their vast explanation of each and every concept present in the book. The RD Sharma class 12th exercise 18.26 deals with the chapter of indefinite integrals which is an essential chapter for performing well in the 12th board exams for every CBSE student. The RD Sharma solutions of Indefinite integrals exercises 18.26 helps you to solve questions efficiently and without much trouble as it lets you experiment with your skills of solving marks and prepares you to perform better for scoring high marks in the board exams and is the subject of.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.26

Indefinite Integrals exercise 18.26 question 1

Answer:
The correct answer is$e^{x} \cos (x)+c$
Hint:
$\frac{d}{d x} \cos x=-\sin x$
Given:
$\int e^{x}(\cos x-\sin x) d x$
Solution:
On integration by part......$\int u \cdot v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]$
$=\int e^{x} \cos x \; d x-\int e^{x} \sin x\; d x$
We know that $\frac{d}{d x} \cos x=-\sin x$
$\begin{aligned} &=\cos x \int e^{x}-\int \frac{d}{d x} \cos x \int e^{x}-\int e^{x} \sin x \; d x \\ &=e^{x} \cos x+\int e^{x} \sin x \; d x-\int e^{x} \sin x\; d x \\ &=e^{x} \cos x+c \end{aligned}$
So the answer is $e^{x} \cos x+c$

Indefinite Integrals exercise 18.26 question 2

Answer:
The correct answer is$\frac{e^{x}}{x^{2}}+c$
Hint: using integration by parts, $\int u . v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]$
Given:
Solution:
$=\int e^{x}\left(\frac{1}{x^{2}}-\frac{2}{x^{3}}\right) d x$
$=\int e^{x} \cdot \frac{1}{x^{2}} d x-2 \int \frac{e^{x}}{x^{3}} d x$
$=\int e^{x} \cdot x^{-2} d x-2 \int \frac{e^{x}}{x^{8}} d x$
$\begin{aligned} &=x^{-2} \int e^{x} d x-\int\left[\frac{d}{d x}\left(x^{-2}\right) \int e^{x} d x\right] d x-2 \int \frac{e^{x}}{x^{3}} d x+c \\ &=x^{-2} \cdot e^{x}-\int-2 x^{-3} e^{x} d x-2 \int \frac{e^{x}}{x^{3}} d x+c \\ &=x^{-2} \cdot e^{x}+c \end{aligned}$
$=\frac{e^{x}}{x^{2}}+c$
So, the correct answer is $\frac{e^{x}}{x^{2}}+c$

Indefinite Integrals exercise 18.26 question 3

Answer:
The correct answer is $e^{x} \tan \frac{x}{2}+c$
Hint:
Using formula:
$\begin{aligned} i.\; \; \; \; &\sin ^{2} x+\cos ^{2} x=1 \\ ii.\; \; \; &\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \end{aligned}$
Given: $\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x$
Solution:
$=\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x$
$=e^{x}\left(\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} 2 \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right) d x$
$=\frac{e^{x}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}}{2\left(\cos \frac{x}{2}\right)^{2}}$
$=\frac{1}{2} e^{x}\left(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}}\right)^{2}$
$\begin{aligned} &=\frac{1}{2} e^{x}\left[\tan \frac{x}{2}+1\right]^{2} \\ &=\frac{1}{2} e^{x}\left[1+\tan \frac{x}{2}\right]^{2} \end{aligned}$
$\begin{aligned} &=\frac{1}{2} e^{x}\left[1+\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right] \\ &=\frac{1}{2} e^{x}\left[\sec ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right] \end{aligned}$
$=e^{x}\left[\frac{1}{2} \sec ^{2} \frac{x}{2}+\tan \frac{x}{2}\right]------(1)$
Suppose $\tan \frac{x}{2}=f(x)$
$f^{\prime}(x)=\frac{1}{2} \sec ^{2} \frac{x}{2}$
We know that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
From equation (1), we get
$\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x=e^{x} \tan \frac{x}{2}+c$
So, the correct answer is $e^{x} \tan \frac{x}{2}+c$

Indefinite Integrals exercise 18.26 question 4

Answer:
The correct answer is$e^{x} \cot x+c$

Given: $\int e^{x} \cdot\left(\cot x-\operatorname{cosec}^{2} x\right) d x$
Solution:
$I=\int e^{x} \cdot\left(\cot x-\operatorname{cosec}^{2} x\right) d x$
On using integration by parts, $\int u . v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]$
$\begin{aligned} &=\int e^{x} \cot x \; d x-\int e^{x} \operatorname{cosec}^{2} x \; d x \\ &=\cot x \int e^{x} d x-\int \frac{d}{d x} \cot x \int e^{x} d x-\int e^{x} \operatorname{cosec}^{2} x\; d x \end{aligned}$
$\begin{aligned} &=\cot x e^{x}+\int e^{x} \operatorname{cosec}^{2} x\; d x-\int e^{x} \operatorname{cosec}^{2} x \; d x \\ &=e^{x} \cot x+c \end{aligned}$
So, the correct answer is $e^{x} \cot x+c$

Indefinite Integrals exercise 18.26 question 5

Answer:
The correct answer is$\frac{e^{x}}{2 x}+c$

Given:
$\int e^{x}\left(\frac{x-1}{2 x^{2}}\right) d x$
Solution:
$I=\int e^{x}\left(\frac{1}{2 x}\right) d x-\int e^{x}\left(\frac{1}{2 x^{2}}\right) d x$
On integrating by parts, $\int u \cdot v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]$
$\begin{aligned} &=\frac{e^{x}}{2 x}-\int e^{x}\left(\frac{d}{d x}\left(\frac{1}{2 x}\right)\right) d x-\int \frac{e^{x}}{2 x} d x \\ &=\frac{e^{x}}{2 x}+\int \frac{e^{x}}{2 x^{2}} d x-\int \frac{e^{x}}{2 x^{2}} d x \\ &=\frac{e^{x}}{2 x}+c \end{aligned}$
So, the correct answer is $\frac{e^{x}}{2 x}+c$

Indefinite Integrals exercise 18.26 question 6

Answer:
The correct answer is$e^{x} \sec x+c$
Hint:
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
$\int e^{x} \sec x(1+\tan x) d x$
Solution:
$I=\int e^{x} \sec x(1+\tan x) d x$
$=\int e^{x}(\sec x+\sec x \tan x) d x$
Put $f(x)=\sec x$
$f^{\prime}(x)=\sec x \tan x$
$\begin{aligned} &=\int e^{x}(\sec x+\sec x \tan x) d x \\ &=e^{x} \sec x+c \end{aligned}$
So, the correct answer is $e^{x} \sec x+c$.

Indefinite Integrals exercise 18.26 question 7

Answer:
The correct answer is$e^{x} \log \sec x+c$

Given:
$\int e^{x}(\tan x-\log \cos x) d x$
Solution:
$I=\int e^{x}(\tan x-\log \cos x) d x$
$=\int e^{x} \tan x\; d x-\int e^{x} \log \cos x \; d x$
Integrating by parts,
$\begin{aligned} &=\int e^{x} \tan x \; d x-e^{x} \log \cos x+\int e^{x}\left(\frac{d}{d x} \log \cos x\right) d x \\ &=\int e^{x} \tan x \; d x-e^{x} \log \cos x-\int e^{x} \tan x \; d x \end{aligned}$
$\begin{aligned} &=-e^{x} \log \cos x+\mathrm{c} \\ &=e^{x} \log \sec x+c \end{aligned}$
So, the correct answer is $e^{x} \log \sec x+c$.

Indefinite Integrals exercise 18.26 question 8

Answer:
The correct answer is$e^{x} \log (\sec x+\tan x)+c$
Hint:
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
$: e^{x}[\sec x+\log (\sec x+\tan x)] d x$
Solution:
$\begin{aligned} &I=\int e^{x}[\sec x+\log (\sec x+\tan x)] d x \\ &f(x)=\log |\sec x+\tan x| \end{aligned}$
$f^{\prime}(x)=\frac{1}{\sec x+\tan x}\left(\sec x \tan x+\sec ^{2} x\right)$
$\begin{aligned} &=\frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \\ &=\sec x \end{aligned}$
$\begin{aligned} &\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c \\ &=e^{x} \log (\sec x+\tan x)+c \end{aligned}$
so, the correct answer is $e^{x} \log (\sec x+\tan x)+c$

Indefinite Integrals exercise 18.26 question 9

Answer:
The correct answer is$e^{x} \log (\sin x)+c$
Hint:
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
Solution:
$\int e^{x}(\cot x+\log \sin x) d x$
$\begin{aligned} &f(x)=\log (\sin x) \\ &f^{\prime}(x)=\frac{1}{\sin x} \cdot \cos x=\cot x \end{aligned}$
$\begin{aligned} &\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c \\ &\int e^{x}\left[(\cot x+\log (\sin x)] d x=e^{x} \log (\sin x)+\mathrm{c}\right. \end{aligned}$
So, the correct answer is $e^{x} \log (\sin x)+c$

Indefinite Integrals exercise 18.26 question 10

Answer:
The correct answer is$\frac{e^{x}}{(x+1)^{2}}+c$
Given:
$\int e^{x} \frac{x-1}{(x+1)^{3}} d x$
Solution:
$I=\int e^{x} \frac{x-1}{(x+1)^{3}} d x$
$\mathrm{I}=\int\left\{\frac{x+1-2}{(x+1)^{3}}\right\} e^{x} d x$
$=\int\left\{\frac{1}{(x+1)^{2}}-\frac{2}{(x+1)^{8}}\right\} e^{x} d x$
$=\int e^{x} \frac{1}{(x+1)^{2}} d x-2 \int e^{x} \frac{1}{(x+1)^{3}} d x$
On using Integration by parts,
$=\left\{\frac{1}{(x+1)^{2}} \cdot e^{x}-\int e^{x} \frac{-2}{(x+1)^{3}} d x\right\}-2 \int e^{x} \frac{1}{(x+1)^{3}} d x$
$=\frac{e^{x}}{(x+1)^{2}}+c$
So, the correct answer $\frac{e^{x}}{(x+1)^{2}}+c$

Indefinite Integrals exercise 18.26 question 11

Answer:
The correct answer is$e^{x} \cot (2 x)+c$
Hint:
$\begin{aligned} i.\; \; \; \; \; &\cos 2 a=1-\sin ^{2} A \\ ii.\; \; \; \; \; &\sin 2 a=2 \sin A \cos A \end{aligned}$
Given: $\int e^{x}\left(\frac{\sin 4 x-4}{1-\cos 4 x}\right) d x$
Solution:
$I=\int e^{x}\left(\frac{\sin 4 x-4}{1-\cos 4 x}\right) d x$
$\mathrm{I}=\int e^{x}\left(\frac{2 \sin (2 x) \cos (2 x)-4}{2 \sin ^{2}(2 x)}\right) d x \quad \therefore\left[\cos 2 a=1-\sin ^{2} A, \sin 2 a=2 \sin A \cos A\right]$
$\mathrm{I}=\int e^{x}\left(\frac{2 \sin (2 x) \cos (2 x)}{2 \sin ^{2}(2 x)}-\frac{4}{2 \sin ^{2}(2 x)}\right) d x$
$\Rightarrow \int e^{x}\left(\cot (2 x)-2 \operatorname{cosec}^{2}(2 x)\right) d x$
$\text { Where } f(x)=\cot (2 x) \text { and } f^{\prime}(x)=-\operatorname{2cosec}^{2}(2 x)$
$\begin{aligned} &I=\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x \\ &=e^{x} f(x)+c \\ &=e^{x} \cot (2 x)+c \end{aligned}$
So, the correct answer is $e^{x} \cot (2 x)+c$

Indefinite Integrals exercise 18.26 question 12

Answer:
The correct answer is$e^{x} \cdot \frac{1}{(1-x)}+c$
Hint:
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
$\int \frac{2-x}{(1-x)^{2}} e^{x} d x$
Solution:
$I=\int \frac{2-x}{(1-x)^{2}} e^{x} d x$
$\begin{aligned} &=\int e^{x}\left(\frac{1+1-x}{(1-x)^{2}}\right) d x \\ &=\int e^{x}\left(\frac{1}{(1-x)^{2}}+\frac{1}{(1-x)}\right) d x \end{aligned}$
we have,
$\int e^{x}\left\{f(x) d x+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
$\begin{aligned} &=\int e^{x}\left(\frac{1}{(1-x)^{2}}+\frac{1}{(1-x)}\right) d x \\ &=e^{x} \cdot \frac{1}{(1-x)}+c \end{aligned}$
So, the correct answer is $e^{x} \cdot \frac{1}{(1-x)}+c$

Indefinite Integrals exercise 18.26 question 13

Answer:
The correct answer is$e^{x} \cdot \frac{1}{x+2}+c$
Hint:
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
$\int e^{x}\left(\frac{1+x}{(2+x)^{2}}\right) d x$
Solution:
$I=\int e^{x}\left(\frac{1+x}{(2+x)^{2}}\right) d x$
$\begin{aligned} &\mathrm{I}=\int e^{x}\left[\frac{x+1+1-1}{(x+2)^{2}}\right] d x \\ &\mathrm{I}=\int e^{x}\left[\frac{x+2}{(x+2)^{2}}-\frac{1}{(x+2)^{2}}\right] d x \\ &\mathrm{I}=\int e^{x}\left[\frac{1}{(x+2)}-\frac{1}{(x+2)^{2}}\right] d x \end{aligned}$
$\text { let } f(x)=\frac{1}{(x+2)} \text { and } f^{\prime}(x)=-\frac{1}{(x+2)^{2}}$
$\begin{aligned} &\mathrm{I}=\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+c \\ &\mathrm{I}=e^{x} \frac{1}{x+2}+c \end{aligned}$

Indefinite Integrals exercise 18.26 question 14

Answer:
The correct answer is$-e^{\frac{-x}{2}} \sec \left(\frac{x}{2}\right)+c$
Hint:
$I=\int e^{x}[f(x)+{f}'(x)]dx=e^{x}f(x)+c$
Given:
$\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{\frac{-x}{2}} d x$
Solution:
$\begin{aligned} &I=\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{\frac{-x}{2}} d x \\ &\operatorname{Let}\left(\frac{-x}{2}\right)=t \quad x=-2 t \\ &\therefore d x=-2 d t \end{aligned}$
$I=\int e^{t} \frac{\sqrt{(1-\sin (-2 t)}}{(1+\cos (-2 t)}(-2dt)$
$=\int e^{t} \frac{\sqrt{(} 1+\sin (2 t)}{(1+\cos (2 t)}(-2 d t)$
$=\int e^{t} \frac{(\cos t+\sin t)}{\cos ^{2} t}(-2 d t)$
$=(-1) \int e^{t}\left\{f(t)+f^{\prime}(t)\right\} d t$
Whose $f(t)=\sec t$ and solution is given by $e^{t} f(t)+c$
$\begin{aligned} &\therefore I=2e^{t} \operatorname{sect}(-1)+c \\ &\therefore I=-2e^{t} \sec t+c \end{aligned}$
So, the correct answer is $-2e^{\frac{-x}{2}} \sec \left(\frac{x}{2}\right)+c$

Indefinite Integrals exercise 18.26 question 15

Answer:
The correct answer is$\log x\; e^{x}+c$
Given:
$\int e^{x}\left(\log x+\frac{1}{x}\right) d x$
Solution:
$I=\int e^{x}\left(\log x+\frac{1}{x}\right) d x$
$=\int e^{x} \log x \; d x+\int e^{x} \cdot \frac{1}{x} d x$
On integration by parts,
$\begin{aligned} &\log x \; e^{x}-\int \frac{1}{x} \cdot e^{x} d x+\int e^{x} \cdot \frac{1}{x} d x \\ &=\log x \; e^{x}+c \end{aligned}$
So, the correct answer is $\log x\; e^{x}+c$

Indefinite Integrals exercise 18.26 question 16

Answer:
The correct answer is$e^{x}(\log x)+e^{x}\left(\frac{-1}{x}\right)+c$
Hint:
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
$\int e^{x}\left(\log x+\frac{1}{x^{2}}\right) d x$
Solution:
$I=\int e^{x}\left(\log x+\frac{1}{x^{2}}\right) d x$
$=\int e^{x}\left(\log x+\frac{1}{x}-\frac{1}{x}+\frac{1}{x^{2}}\right) d x$
$\int e^{x}\left(\log x+\frac{1}{x}\right) d x+\int e^{x}\left(-\frac{1}{x}+\frac{1}{x^{2}}\right) d x$
We have,
$\begin{aligned} &\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c \\ &\int e^{x}\left(\log x+\frac{1}{x}\right) d x+\int e^{x}\left(-\frac{1}{x}+\frac{1}{x^{2}}\right) d x \end{aligned}$
$=e^{x}(\log x)+e^{x}\left(\frac{-1}{x}\right)+c$
So, the correct answer is $e^{x}(\log x)+e^{x}\left(\frac{-1}{x}\right)+c$

Indefinite Integrals exercise 18.26 question 17

Answer:
The correct answer is $e^{x}(\log x)^{2}+c$
Hint:
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
$\int \frac{e^{x}}{x}\left\{x(\log x)^{2}+2 \log x\right\} d x$
Solution:
$I=\int \frac{e^{x}}{x}\left\{x(\log x)^{2}+2 \log x\right\} d x$
$\int e^{x}\left\{(\log x)^{2}+2 \log x \cdot \frac{1}{x}\right\} d x$
We have, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
$\text { Let } f(x)=(\log x)^{2}, f^{\prime}(x)=2 \log x \cdot \frac{1}{x}$
$\begin{aligned} &f(x)+f^{\prime}(x)=(\log x)^{2}+2 \log x \cdot \frac{1}{x} \\ &I=\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x \quad \text { if } f(x)=(\log x)^{2} \end{aligned}$
$\begin{aligned} &=e^{x} f(x)+c \\ &=e^{x}(\log x)^{2}+c \end{aligned}$
So, the correct answer is $e^{x}(\log x)^{2}+c$

Indefinite Integrals exercise 18.26 question 18

Answer:
The correct answer is$e^{x} \sin ^{-1} x+c$
Hint:
Using ILATE rule
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
Given:
$\int e^{x} \frac{\sqrt{1-x^{2}} \sin ^{-1} x+1}{\sqrt{1-x^{2}}} d x$
Solution:
$I=\int e^{x} \frac{\sqrt{1-x^{2}} \sin ^{-1} x+1}{\sqrt{1-x^{2}}} d x$
It is in the form of,
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x$
Where $f(x)=\sin ^{-1} x$
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c$
So,
$\int e^{x}\left(\frac{1}{\sqrt{1-x^{2}}}+\sin ^{-1} x\right)=e^{x} \sin ^{-1} x+c$
So the correct answer is $e^{x} \sin ^{-1} x+c$

Indefinite Integrals exercise 18.26 question 19

Answer:
The correct answer is$e^{2 x} \cos x+c$

Given:
$\int e^{2 x}(-\sin x+2 \cos x) d x$
Solution:
$I=\int e^{2 x}(-\sin x+2 \cos x) d x$
$=-\int e^{2 x} \sin x \; d x+2 \int e^{2 x} \cos x\; d x$
Consider $2 \int e^{2 x} \cos x \; d x$
Integrating by parts, we get
$\begin{aligned} &\text { Let } u=\cos x \\ &d u=-\sin x \; d x \\ &v=e^{2 x} \end{aligned}$
$\int v d x=\frac{e^{2 x}}{2}$
$2 \int e^{2 x} \cos x d x$
$\begin{aligned} &=2\left[\int \frac{e^{2 x}}{2} \cos x-\int \frac{e^{2 x}}{2}(-\sin x) d x\right]+c \\ &=e^{2 x} \cos x+\int e^{2 x} \sin x \; d x+c \end{aligned}$
$\text { Now, } I=-\int e^{2 x} \sin x d x+2 \int e^{2 x} \cos x d x$
$\begin{aligned} &=-\int e^{2 x} \sin x \; d x+e^{2 x} \cos x+\int e^{2 x} \sin x \; d x+c \\ &=e^{2 x} \cos x+c \end{aligned}$
So, the correct answer is $e^{2 x} \cos x+c$

Indefinite Integrals exercise 18.26 question 20

Answer:
The correct answer is$e^{x} \tan ^{-1} x+c$
Hint:
$\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$
Given:
$\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$
Solution:
$\begin{aligned} &\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x \\ &\text { Let } f(x)=\tan ^{-1} x \\ &\text { Then } f^{\prime}(x)=\frac{1}{1+x^{2}} \end{aligned}$
$\begin{aligned} &\text { Now, } I=\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x \\ &I=e^{x} f(x)+c \\ &I=e^{x} \tan ^{-1} x+c \end{aligned}$
So the correct answer is $e^{x} \tan ^{-1} x+c$

Indefinite Integrals exercise 18.26 question 21

Answer:
The correct answer is$e^{x} \cot x+c$

Given:
$\int e^{x} \frac{\sin x \cos x-1}{\sin ^{2} x} d x$
Solution:
$\begin{aligned} &I=\int e^{x} \frac{\sin x \cos x-1}{\sin ^{2} x} d x \\ &I=\int e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right) d x \\ &I=\int e^{x} \cot x d x-\int e^{x} \operatorname{cosec}^{2} x \; d x \end{aligned}$
Consider $\int e^{x} \operatorname{cosec}^{2} x\; d x$

$\int e^{x} \operatorname{cosec}^{2} x \; d x=-e^{x} \cot x+\int e^{x} \cot x\; d x$
$\begin{aligned} &I=\int e^{x} \cot x \; d x-\int e^{x} \operatorname{cosec}^{2} x \; d x \\ &=\int e^{x} \cot x \; d x+e^{x} \cot x-\int e^{x} \cot x \; d x+c \\ &=e^{x} \cot x+c \end{aligned}$
So, the correct answer is $e^{x} \cot x+c$

Indefinite Integrals exercise 18.26 question 22

Answer:
The correct answer is$x \tan (\log x)+c$
Hint:
$\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$
Given:
$\int\left[\tan (\log x)+\sec ^{2}(\log x)\right] d x$
Solution:
$I=\int\left[\tan (\log x)+\sec ^{2}(\log x)\right] d x$
$\begin{aligned} &\text { Put } \log x=t \\ &\Rightarrow x=e^{t} \\ &d x=e^{t} d t \end{aligned}$
$I=\int\left(\tan t+\sec ^{2} t\right) e^{t} d t$
$\begin{aligned} &\text { Here, } f(t)=\tan t \\ &f^{\prime}(t)=\sec ^{2} t \\ &\text { Let } e^{t} \tan t=p \end{aligned}$
Differentiate both sides w.r.t ‘t’
$\begin{aligned} &e^{t}\left(\tan t+\sec ^{2} t\right)=\frac{d p}{d t} \\ &e^{t}\left(\tan t+\sec ^{2} t\right) d t=d p \end{aligned}$
$\begin{aligned} &I=\int d p \\ &=p+c \\ &=e^{t} \tan t+c \\ &=x \tan (\log x)+c \end{aligned}$
So, the correct answer is $x \tan (\log x)+c$

Indefinite Integrals exercise 18.26 question 23

Answer:
The correct answer is$e^{x} \frac{1}{(x-2)^{2}}+c$
Hint:
$\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$
Given:
$\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x$
Solution:
$\begin{aligned} &I=\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x \\ &=\int e^{x} \frac{(x-2-2)}{(x-2)^{3}} d x \\ &=\int e^{x}\left[\frac{(x-2)}{(x-2)^{3}}-\frac{2}{(x-2)^{3}}\right] d x \end{aligned}$
$=\int e^{x}\left[\frac{1}{(x-2)^{2}}+\frac{-2}{(x-2)^{3}}\right] d x$
Now, consider,
$f(x)=\frac{1}{(x-2)^{2}}$
Then $f^{\prime}(x)=\frac{-2}{(x-2)^{3}}$
Thus the above integral is of the form
$e^{x}\left(f(x)+f^{\prime}(x)\right)$
As,
$\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$
$I=e^{x} \frac{1}{(x-2)^{2}}+c$
So, the correct answer is $e^{x} \frac{1}{(x-2)^{2}}+c$

Indefinite Integrals exercise 18.26 question 24

Answer:
The correct answer is$-\frac{1}{2} e^{2 x} \cot x+c$
Hint:
$\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c$
Given:
$\int e^{2 x} \frac{(1-\sin 2 x)}{(1-\cos 2 x)} d x$
Solution:
$I=\int e^{2 x} \frac{(1-\sin 2 x)}{(1-\cos 2 x)} d x$
$\begin{aligned} &\text { Let } 2 x=t \\ &\operatorname{Or} 2 d x=d t \end{aligned}$
$\begin{aligned} &I=\int e^{t} \frac{(1-\sin t)}{(1-\cos t)} \frac{d t}{2} \\ &=\frac{1}{2} \int e^{t} \frac{\left(1-2 \sin \frac{t}{2} \cos \frac{t}{2}\right)}{2 \sin ^{2} \frac{t}{2}} d t \end{aligned}$
$\begin{aligned} &=\frac{1}{4} \int e^{t}\left(\operatorname{cosec}^{2} \frac{t}{2}-2 \cot \frac{t}{2}\right) d t \\ &=\frac{1}{4} \int e^{t}\left(-2 \cot \frac{t}{2}+\operatorname{cosec}^{2} \frac{t}{2}\right) d t \end{aligned}$
$\begin{aligned} &\text { Let } f(x)=-2 \cot \frac{t}{2} \\ &\text { Or } f^{\prime}(x)=\operatorname{cosec}^{2} \frac{t}{2} d t \end{aligned}$
$\left[\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c\right]$
$\begin{aligned} &\therefore \quad I=\frac{1}{4} e^{t}\left(-2 \cot \frac{t}{2}+c\right) \\ &\operatorname{Or} I=-\frac{1}{2} e^{2 x} \cot x+c \end{aligned}$
So, the correct answer is $-\frac{1}{2} e^{2 x} \cot x+c$

RD Sharma class 12th exercise 18.26 contains 24 questions from the Class 12 RD Sharma chapter 18 exercise 18.26 solution provides with the best simple material to understand the important concepts of the chapter like, formula of fundamental integration, special integration formula, integration of trigno by parts, important integral theorem.

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