RD Sharma Class 12 Exercise 18.26 Indefinite integrals Solutions Maths

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

Edited By Kuldeep Maurya | Updated on Jan 24, 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.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

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.

Exercises provided in the RD Sharma class 12 chapter 18 exercise 18.26 are designed by professionals who are experts in maths. The experts not only provide you with extraordinary exercises but also helpful advice and tips so that you can solve the questions in an alternate way which saves your time and your hard work. The RD Sharma class 12 chapter 18 exercise 18.26 matches up with the syllabus of NCERT that allows it to be one of the most beneficial solutions for students appearing for the public examinations.

The RD Sharma class 12th exercise 18.26 ensures that every student gets appropriate knowledge about every concept and therefore they pass on the understanding of the chapter through better results in the board exams. RD Sharma class 12 solutions chapter 18 exercise 18.26 are not only helpful in better performance for exams but also helps you to be outstanding in your class by staying ahead of the class because starting from R D Sharma solutions can help you stay ahead of the class as the teachers make use of the solutions to provide Nexus and also to assign homework.

The RD Sharma class 12th exercise 18.26 can easily be downloaded from the Careers360 website and any student can make use of the available PDF and online study materials from the website which doesn't cause them any amount to make use of these solutions

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. What is the number of exercises given in the RD Sharma solution for class 12?

There are a total of 36 exercises to give you a brief understanding of the concepts given in the chapter clearly

2. Can I solve homework by using the R D Sharma solutions?

Yes, you can definitely use the RD Sharma solutions to solve homework more efficiently and in less time.

3. Is the indefinite integrals an important chapter for 12th board exams?

Every chapter mentioned in the RD Sharma solutions are important for your preparation and to score good marks in the Board examinations.

4. Is RD Sharma better than NCERT?

Yes, RD Sharma is better than NCERT in every way possible, as the RD Sharma solutions give you solved questions and brief exercises to practice thoroughly in comparison to the NCERT textbooks.

5. Can I buy the RD Sharma solutions from offline stores?

You don't need to trouble Yourself by visiting stores to buy the offline versions of the RD Sharma solution when you can easily download it from the careers360 website for free of cost.