RD Sharma Class 12 Exercise 18.23 Indefinite integrals Solutions Maths

RD Sharma Class 12 Exercise 18.23 Indefinite integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:26 PM IST

The Class 12 RD Sharma chapter 18 exercise 18.23 solution deals with the chapter of indefinite integrals and covers up some of the concepts of the chapter. As this is very vast and complex as well, it takes a lot of time for students to complete this chapter, that is why RD Sharma class 12 solution of Indefinite integrals exercise 18.23 RD Sharma Solutions has been recommended by schools to be used by students as it divides the chapter into equal portions and with better concepts to make it basic and simple for students to understand.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise: 18.23
Benefits of using the RD Sharma class 12th exercise 18.23 are given below:-

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise: 18.23

Indefinite Integrals Exercise 18.23 Question 1

Answer : $\frac{2}{3} \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)+c$
Hint: To solve this equation we have to differentiate it and change cos in terms of tan
Given : $\int \frac{1}{5+4 \cos x} d x$
Solution : $\frac{1}{5+4 \cos x}$
Use the formula : $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$=\frac{1}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)}$
$\begin{aligned} &=\frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} \\ &=\frac{1+\tan ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \\ &=\frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} \\ &I=\int \frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} d x \end{aligned}$
Let,
$\begin{aligned} &\tan \frac{x}{2}=t \\ &\frac{1}{2} \sec ^{2} \frac{x}{2}=\frac{d t}{d x} \\ &d x=\frac{d t}{\frac{1}{2} \sec ^{2} \frac{x}{2}} \\ &I=\int \frac{\sec ^{2} \frac{x}{2}}{9+t^{2}} \cdot\frac{d t}{\frac{1}{2} \sec ^{2} \frac{x}{2}} \end{aligned}$
$\begin{aligned} &I=\int \frac{2 d t}{9+t^{2}} \\ &I=\frac{2}{3} \tan ^{-1}\left(\frac{t}{3}\right)+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right] \\ &I=\frac{2}{3} \tan ^{-1}\left[\frac{\tan \frac{x}{2}}{3}\right]+c \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 2

Answer : $\frac{2}{3} \tan ^{-1}\left[\frac{5 \tan \frac{x}{2}-4}{3}\right]+c$
Hint: To solve this equation we have to use $\sin ^{2} x+\cos ^{2} x=1$ and divide numerator and denominator by $\cos ^{2} \frac{x}{2}$
Given : $\int \frac{1}{5-4 \sin x} d x$
Solution :
$\begin{aligned} &I=\int \frac{1}{5\left(\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}\right)-4\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} d x \; \; \; \; \; \; \; \; \; \; \quad\left[\sin ^{2} x+\cos ^{2} x=1, \sin 2 x=2 \sin x \cos x\right] \\ &=\int \frac{d x}{5 \sin ^{2} \frac{x}{2}+5 \cos ^{2} \frac{x}{2}-8 \sin \frac{x}{2} \cos \frac{x}{2}} \end{aligned}$
Divide numerator and denominator by $\cos ^{2} \frac{x}{2}$
$I=\int \frac{\sec ^{2} \frac{x}{2}}{5 \tan ^{2} \frac{x}{2}+5-8 \tan \frac{x}{2}} d x$
Let
$\begin{aligned} &\tan \frac{x}{2}=t \\ &\frac{\sec ^{2} \frac{x}{2}}{2} d x=d t \\ &I=2 \int \frac{1}{5 t^{2}+5-8 t} d t \\ &I=2 \times \frac{1}{5} \int \frac{1}{t^{2}-\frac{8 t}{5}+1} d t \end{aligned}$
$\begin{aligned} &I=\frac{2}{5} \int \frac{d t}{t^{2}-\frac{8 t}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} \\ &I=\frac{2}{5} \int \frac{d t}{\left(t-\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} \end{aligned}$
$\begin{aligned} &I=\frac{2}{5} \times \frac{5}{3} \tan ^{-1}\left|\frac{t-\frac{4}{5}}{\frac{3}{5}}\right|+c \\ &I=\frac{2}{3} \tan ^{-1}\left[\frac{5 t-4}{3}\right]+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right] \\ &I=\frac{2}{3} \tan ^{-1}\left[\frac{5 \tan \frac{x}{2}-4}{3}\right]+c \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 3

Answer : $\frac{1}{\sqrt{3}} \log \left|\frac{\tan \frac{x}{2}-2-\sqrt{3}}{\tan \frac{x}{2}-2+\sqrt{3}}\right|+c$
Hint : To solve this equation we have to convert sin into tan form
Given : $\int \frac{1}{1-2 \sin x} d x$
Solution : $I=\int \frac{1}{1-2 \sin x} d x$
Using the formula : $\left[\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right]$
$\begin{aligned} &I=\int \frac{1}{1-\frac{4 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}-4 \tan \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{\left[\tan \frac{x}{2}-2\right]^{2}-4+1} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{\left[\tan \frac{x}{2}-2\right]^{2}-(\sqrt{3})^{2}} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{\sec ^{2} \frac{x}{2}}{\left[\tan \frac{x}{2}-2\right]^{2}-(\sqrt{3})^{2}} d x \\ &{\left[\begin{array}{l} t=\tan \frac{x}{2} \\ d t=\sec ^{2} \frac{x}{2} \times \frac{1}{2} d x \\ 2 d t=\left(\sec ^{2} \frac{x}{2}\right) d x \end{array}\right]} \end{aligned}$
$\begin{aligned} &I=\int \frac{2 d t}{(t-2)^{2}-(\sqrt{3})^{2}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right] \\ &I=\frac{2}{2 \sqrt{3}} \log \left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 4

Answer : $\frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5} \tan \frac{x}{2}}{\sqrt{3}-\sqrt{5} \tan \frac{x}{2}}\right|+c$
Hint: To solve this statement we have to convert cos into tan form
Given : $\int \frac{1}{4 \cos x-1} d x$
Solution :
$\int \frac{1}{4 \cos x-1} d x$
Use the formula : $\left[\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right]$
$\begin{aligned} &=\int \frac{1}{4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)-1} d x \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{4-4 \tan ^{2} \frac{x}{2}-1-\tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{3-5 \tan ^{2} \frac{x}{2}} d x \end{aligned}$
$\begin{aligned} &{\left[\begin{array}{l} \tan \frac{x}{2}=t \\ \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t \end{array}\right]} \\ &=\int \frac{2 d t}{3-5 t^{2}} \\ &=\frac{1}{\sqrt{3}} \int \frac{2 \sqrt{3}}{(\sqrt{3}-\sqrt{5} t)(\sqrt{3}+\sqrt{5} t)} d t \\ &=\frac{1}{\sqrt{3}} \int \frac{(\sqrt{3}-\sqrt{5} t)+(\sqrt{3}+\sqrt{5} t)}{(\sqrt{3}-\sqrt{5} t)(\sqrt{3}+\sqrt{5} t)} d t \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{3}} \int \frac{d t}{\sqrt{3}+\sqrt{5} t}+\frac{1}{\sqrt{3}} \int \frac{d t}{\sqrt{3}-\sqrt{5} t}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{a t+b}=\log \left|\frac{a t+b}{a}\right|+c\right] \\ &=\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{3}+\sqrt{5} t}{\sqrt{5}}\right|+\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{3}-\sqrt{5} t}{-\sqrt{5}}\right|+c \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{3}+\sqrt{5} t}{\sqrt{3}-\sqrt{5} t}\right|+c \\ &=\frac{1}{\sqrt{15}} \log \left|\frac{\sqrt{3}+\sqrt{5} \tan \frac{x}{2}}{\sqrt{3}-\sqrt{5} \tan \frac{x}{2}}\right|+C \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 5

Answer : $-\log \left|1-\tan \frac{x}{2}\right|+c$
Hint: To solve this question we have to convert sinx and cosx in terms of tanx
Given: $\int \frac{1}{1-\sin x+\cos x} d x$
Solution : $\int \frac{1}{1-\sin x+\cos x} d x$
$\left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]$
$\begin{aligned} &I=\int \frac{1}{1-\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}-2 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{2-2 \tan \frac{x}{2}} d x \\ &I=\int \frac{\sec ^{2} \frac{x}{2}}{2\left(1-\tan \frac{x}{2}\right)} d x \end{aligned}$
$\begin{aligned} &{\left[\begin{array}{l} t=\tan \frac{x}{2} \\ \frac{d t}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \end{array}\right]} \\ &\begin{array}{l} I=\int \frac{d t}{1-t} \\ {\left[\begin{array}{l} 1-t=u \\ -1=\frac{d u}{d t} \\ d t=-d u \end{array}\right]} \end{array} \end{aligned}$
$\begin{aligned} &\left.I=\int-\frac{d u}{u}=-\log |u|+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { [ } \int \frac{d x}{x}=\log |x|+c\right] \\ &I=-\log |1-t|+c \\ &I=-\log \left|1-\tan \frac{x}{2}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 6

Answer : $\tan ^{-1}\left[1+\tan \frac{x}{2}\right]+c$
Hint: To solve this statement we have to convert sinx and cosx in terms of tanx.
Given: $\int \frac{1}{3+2 \sin x+\cos x} d x$
Solution : $I=\int \frac{1}{3+2 \sin x+\cos x} d x$
We use the formula : $\left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]$
$\begin{aligned} &I=\int \frac{1}{3+2 \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{3+3 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{\sec ^{2} \frac{x}{2}}{2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+4} d x \\ &I=\int \frac{\sec ^{2} \frac{x}{2}}{2\left(\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+2\right)} d x \end{aligned}$
$\begin{aligned} &{\left[\begin{array}{l} t=\tan \frac{x}{2} \\ \frac{d t}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \end{array}\right]} \\ &I=\int \frac{d t}{t^{2}+2 t+2} \\ &I=\int \frac{d t}{\left(t^{2}+2 t+1\right)+1} \end{aligned}$
$\begin{aligned} &I=\int \frac{d t}{(t+1)^{2}+1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right] \\ &I=\frac{1}{1} \tan ^{-1}(t+1)+c \\ &I=\tan ^{-1}\left(\tan \frac{x}{2}+1\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 7

Answer : $\frac{1}{6} \tan ^{-1}\left[\frac{5 \tan \frac{x}{2}+2}{6}\right]+c$
Hint: To solve this equation we have to convert cosx and sinx in terms of tanx.
Given: $\int \frac{1}{13+3 \cos x+4 \sin x} d x$
Solution : $I=\int \frac{1}{13+3 \cos x+4 \sin x} d x$
We will use the formula : $\left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]$
$\begin{aligned} &I=\int \frac{1}{13+3\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+4\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{13+13 \tan ^{2} \frac{x}{2}+3-3 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \end{aligned}$
$I=\int \frac{\sec ^{2} \frac{x}{2}}{10 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}+16} d x$
$\begin{aligned} &=\int \frac{\sec ^{2} \frac{x}{2}}{2\left(5 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+8\right)} d x \\ &\text { put } t=\tan \frac{x}{2} \\ &\frac{d t}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \\ &I=\int \frac{d t}{\left(5 t^{2}+4 t+8\right)} \end{aligned}$
$\begin{aligned} &=\int \frac{d t}{5\left(t^{2}+\frac{4}{5}+\frac{8}{5}\right)} \\ &=\int \frac{d t}{5\left(t^{2}+2 \frac{2}{5} t+\left(\frac{2}{5}\right)^{2}-\left(\frac{2}{5}\right)^{2}+\frac{8}{5}\right)} \\ &=\frac{1}{5} \int \frac{d t}{\left(t+\frac{2}{5}\right)^{2}+\frac{36}{25}} \end{aligned}$
$=\frac{1}{5} \int \frac{d u}{\left(t+\frac{2}{5}\right)^{2}+\left(\frac{6}{5}\right)^{2}}$
Use the formula : $\left[\int\left(\frac{d x}{x^{2}+a^{2}}\right)=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\right]$
$\begin{aligned} &I=\frac{1}{5} \times \frac{5}{6} \tan ^{-1}\left(\frac{\left(t+\frac{2}{5}\right)}{\frac{6}{5}}\right)+c \\ &I=\frac{1}{6} \tan ^{-1}\left(\frac{\frac{5 t+2}{5}}{\frac{6}{5}}\right)+c \\ &I=\frac{1}{6} \tan ^{-1}\left(\frac{5 \tan \frac{x}{2}+2}{6}\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 8

Answer : $\mathrm{I}=\frac{-1}{\sqrt{2}} \log \left|\cos e c\left(x-\frac{\pi}{4}\right)+\cot \left(x-\frac{\pi}{4}\right)\right|+c$
Hint: To solve this equation we have to use the formula of $\cos A-\sin B$
Given: $\int \frac{1}{\cos x-\sin x} d x$
Solution :
$\begin{aligned} &\int \frac{(\sqrt{2})}{\sqrt{2}(\cos x-\sin x)} d x \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \sin x} d x \\ &=\frac{1}{\sqrt{2}} \int \frac{1}{\sin \frac{\pi}{4} \cos x-\sin x \cos \frac{\pi}{4}} d x \end{aligned}$
$\begin{aligned} &=\frac{1}{\sqrt{2}} \int \frac{1}{\sin \left(\frac{\pi}{4}-x\right)} d x \; \; \; \; \; \; \; \; \; \; \; \quad[\sin a \cos b-\cos a \sin b=\sin (a-b)] \\ &=\frac{-1}{\sqrt{2}} \int \frac{1}{\sin \left(x-\frac{\pi}{4}\right)} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; [\sin (-x)=-\sin x] \end{aligned}$
$\begin{aligned} &=\frac{-1}{\sqrt{2}} \int \cos e c\left(x-\frac{\pi}{4}\right) d x \\ &=\frac{-1}{\sqrt{2}} \log \left|\cos e c\left(x-\frac{\pi}{4}\right)+\cot \left(x-\frac{\pi}{4}\right)\right|+c \end{aligned}$
Note : Final answer is not matching.

Indefinite Integrals Exercise 18.23 Question 9

Answer : $\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\tan x / 2-1}{\sqrt{2}+\tan x / 2+1}\right|+c$
Hint : To solve this equation we use formula of $\cos (A-B)$
Given : $\int \frac{1}{\sin x+\cos x} d x$
Solution : Let $I=\int \frac{1}{\sin x+\cos x} d x$
$\begin{aligned} &=\frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2}} \\ &=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right) \\ &=\sqrt{2}\left(\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cos x\right) \end{aligned}$
$\begin{aligned} &=\sqrt{2}\left(\cos x \cdot \cos \frac{\pi}{4}+\sin x \sin \frac{\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\cos (A-B)=\cos A \cos B+\sin A \sin B] \\ &I=\sqrt{2}\left[\cos \left(x-\frac{\pi}{4}\right)\right] \\ &I=\int \frac{d x}{\sin x+\cos x}=\frac{1}{\sqrt{2}} \int \frac{d x}{\cos \left(x-\frac{\pi}{4}\right)} \\ &I=\frac{1}{\sqrt{2}} \int \sec \left(x-\frac{\pi}{4}\right) d x \end{aligned}$
$\begin{aligned} &{\left[\int \sec x d x=\log |\sec x+\tan x|+c\right]} \\ &=\frac{1}{\sqrt{2}} \log \left|\sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right)\right|+c \end{aligned}$
Note : Final answer is not matching.

Indefinite Integrals Exercise 18.23 Question 10

Answer : $\frac{2}{3} \tan ^{-1}\left[3 \tan \frac{x}{2}\right]+c$
Hint: To solve this equation we have to change cosx into tanx form
Given: $\int \frac{1}{5-4 \cos x} d x$
Solution : $\int \frac{1}{5-4 \cos x}dx$
$\operatorname{Cos} x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$\begin{aligned} &=\int \frac{1}{5-4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} \mathrm{d} x \\ &= \\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{1+9 \tan ^{2} \frac{x}{2}} d x \end{aligned}$
$\begin{aligned} &=\int \frac{\sec ^{2} \frac{x}{2}}{1+9 \tan ^{2} \frac{x}{2}} d x \\ &=\int \frac{\sec ^{2} \frac{x}{2}}{1+\left(3 \tan \frac{x}{2}\right)^{2}} d x \end{aligned}$
$\begin{aligned} &\text { Let } 3 \tan \frac{x}{2}=t \\ &=3 \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x=d t \end{aligned}$
$\begin{aligned} &=\frac{2}{3} \int \frac{d t}{1+t^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\langle\int\frac{1}{a^{2}+t^{2}}dt=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)\right\rangle \\ &=\frac{2}{3} \tan ^{-1}(t)+C \\ &=\frac{2}{3} \tan ^{-1}\left(3 \tan \frac{x}{2}\right)+C \end{aligned}$

Indefinite Integrals Exercise 18.23 Question 11

Answer : $\sqrt{2} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)+1}{\sqrt{2}}\right)+c$
Hint : To solve this equation we have to convert sinx and cosx in terms of tanx
Given : $\int \frac{1}{2+\sin x+\cos x} d x$
Solution : We use the formula : $\left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]$
$\begin{aligned} &=\int \frac{1}{2+\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{2+2 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \end{aligned}$
Multiple 2 by numerators and denominator
$\int \frac{2\left(\sec ^{2} \frac{x}{2}\right)}{2\left(\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+3\right)} d x$
$\begin{aligned} &t=\tan \frac{x}{2} \\ &\frac{d t}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \\ &d t=\frac{1}{2} \sec ^{2} \frac{x}{2} d x \end{aligned}$
$\begin{aligned} &=2 \int \frac{d t}{t^{2}+2 t+3} \\ &=2 \int \frac{d t}{\left(t^{2}+2 t+1\right)+2} \\ &=2 \int \frac{d t}{(t+1)^{2}+2} \end{aligned}$
$\begin{aligned} &t+1=u \\ &1=\frac{d u}{d t} \\ &d u=d t \\ &=2 \int \frac{d u}{(u)^{2}+(\sqrt{2})^{2}} \end{aligned}$
$\begin{aligned} &\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ &=\sqrt{2} \times \frac{\sqrt{2}}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+c \\ &=\sqrt{2} \tan ^{-1}\left(\frac{t+1}{\sqrt{2}}\right)+c \end{aligned}$
$=\sqrt{2} \tan ^{-1}\left(\frac{\tan \frac{x}{2}+1}{\sqrt{2}}\right)+c$

Indefinite Integrals Exercise 18.23 Question 12

Answer : $I=-\frac{1}{2} \log \left|\cos e c\left(x+\frac{\pi}{3}\right)+c \operatorname{ot}\left(x+\frac{\pi}{3}\right)\right|$
Hint: To solve this question we have to rewrite the formula of cosx and sinx by adding the terms
$\sin \frac{\pi}{3} \operatorname{and} \cos \frac{\pi}{3}$
Given : $I=\int \frac{1}{\sin x+\sqrt{3} \cos x} d x$
Solution :
$\begin{aligned} &I=\int \frac{1}{\sin x+\sqrt{3} \cos x} d x \\ &\sin x+\sqrt{3} \cos x=2\left[\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x\right] \\ &=2\left[\cos \frac{\pi}{3} \sin x+\sin \frac{\pi}{3} \cos x\right] \\ &=2\left[\sin \left(x+\frac{\pi}{3}\right)\right] \end{aligned}$
$\begin{aligned} &I=\int \frac{1}{2 \sin \left(x+\frac{\pi}{3}\right)} d x \\ &I=\frac{1}{2} \int \cos e c\left(x+\frac{\pi}{3}\right) d x \\ &\int(\cos \operatorname{ecxd} x)=-\log |\cos e c x+\cot x|+c \end{aligned}$
$I=-\frac{1}{2} \log \left|\operatorname{cosec}\left(x+\frac{\pi}{3}\right)+\cot \left(x+\frac{\pi}{3}\right)\right|+c$
Note : Final answer is not matching.

Indefinite Integrals Exercise 18.23 Question 13

Answer : $\frac{1}{2} \log \left|\sec \left(x-\frac{\pi}{3}\right)+\tan \left(x-\frac{\pi}{3}\right)\right|+c$
Hint: To solve this question we have to use formula of $\cos \left(x-\frac{\pi}{3}\right)$
Given: $\int \frac{1}{\sqrt{3} \sin x+\cos x} d x$
Solution : Multiplying and dividing the denominator by 2
$\begin{aligned} &=\int \frac{1}{2\left(\frac{\sqrt{3}}{2} \sin x+\left(\frac{1}{2}\right) \cos x\right)} d x \\ &=\frac{1}{2} \int \frac{1}{\frac{1}{2} \cos x+\frac{\sqrt{3}}{2} \sin x} d x \\ &=\frac{1}{2} \int \frac{1}{\cos \frac{\pi}{3} \cos x+\sin \frac{\pi}{3} \sin x} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \cos (\mathrm{A}-\mathrm{B})=\cos A \cos \mathrm{B}+\sin A \sin \mathrm{s} \\ &=\frac{1}{2} \int \frac{1}{\cos \left(x-\frac{\pi}{3}\right)} d x \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int \sec \left(x-\frac{\pi}{3}\right) d x \\ &t=x-\frac{\pi}{3} \\ &\frac{d t}{d x}=1 \\ &d t=d x \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int \sec t d t \\ &=\frac{1}{2} \log |\sec t+\tan t|+c \\ &=\frac{1}{2} \log \left|\sec \left(x-\frac{\pi}{3}\right)+\tan \left(x-\frac{\pi}{3}\right)\right|+c \end{aligned}$
Note : Final answer is not matching with the book.

Indefinite Integrals Exercise 18.23 Question 14

Answer : $=-\frac{1}{2} \log \left|\operatorname{cosec}\left(x-\frac{\pi}{3}\right)+\cot \left(x-\frac{\pi}{3}\right)\right|+c$
Hint : To solve this question we have to use formula of $\sin \left(x-\frac{\pi}{3}\right)$
Given : $\int \frac{1}{\sin x-\sqrt{3} \cos x} d x$
Solution : $\int \frac{1}{\sin x-\sqrt{3} \cos x} dx$
$\begin{aligned} &=\int \frac{1}{2\left(\frac{1}{2} \sin x-\frac{\sqrt{3}}{2} \cos x\right)} d x \\ &=\frac{1}{2} \int \frac{1}{\sin x \cos \frac{\pi}{3}-\cos x \sin \frac{\pi}{3}} d x \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int \frac{1}{\sin \left(x-\frac{\pi}{3}\right)} d x \\ &=\frac{1}{2} \int \operatorname{cosec}\left(x-\frac{\pi}{3}\right) d x \\ &t=\left(x-\frac{\pi}{3}\right) \end{aligned}$
$\begin{aligned} &\frac{d t}{d x}=1 \\ &d t=d x \\ &=-\frac{1}{2} \log |\cos e c t+\cos t|+c \\ &=-\frac{1}{2} \log \left|\cos e c\left(x-\frac{\pi}{3}\right)+\cot \left(x-\frac{\pi}{3}\right)\right|+c \end{aligned}$
Note: Final answer is not matching with the book.

Indefinite Integrals Exercise 18.23 Question 15

Answer : $\frac{1}{5} \log \left|\frac{\tan \frac{x}{2}+2}{\tan \frac{x}{2}-3}\right|+C$
Hint: To solve this question we have to use partial equation
Given: $\int \frac{1}{5+7 \cos x+\sin x} d x$
Solution :
$I=\int \frac{1}{5+7\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)+\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$ $\sin 2 \theta=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \cos 2 \theta=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$\begin{aligned} &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+7\left(1-\tan ^{2} \frac{x}{2}\right)+2 \tan \frac{x}{2}} d x \\ &I=\int \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+7-7 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}} d x \end{aligned}$
$\begin{aligned} &I=\int \frac{\sec ^{2} \frac{x}{2}}{12-2 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}} d x\\ &=\frac{1}{2} \int \frac{\sec ^{2} \frac{x}{2}}{6-\tan ^{2} \frac{x}{2}+\tan \frac{x}{2}} d x \end{aligned}$
$\begin{aligned} &\text { Put } \tan \frac{x}{2}=t \\ &\ \sec ^{2} \frac{x}{2}\cdot\frac{1}{2} d x=d t \\ &I=\int \frac{d t}{6-t^{2}+t} \\ &=\int \frac{-d t}{t^{2}-t-6} \\ &=\int \frac{-d t}{t^{2}-3 t+2 t-6} \end{aligned}$
$\begin{aligned} &=\int \frac{-d t}{t(t-3)+2(t-3)} \\ &=-\frac{1}{5} \int \frac{(3+2) d t}{(t-3)(t+2)} \\ &=-\frac{1}{5} \int \frac{[(t+2)-(t-3)] d t}{(t-3)(t+2)} \\ &=-\frac{1}{5} \int \frac{d t}{(t-3)}+\frac{1}{5} \int \frac{d t}{(t+2)} \end{aligned}$ $\begin{aligned} &=\frac{1}{5}[\log |t+2|-\log |t-3|]+C \\ &=\frac{1}{5} \log \left|\frac{t+2}{t-3}\right|+C \\ &=\frac{1}{5} \log \left|\frac{\tan \frac{x}{2}+2}{\tan \frac{x}{2}-3}\right|+C \end{aligned}$

The exercise in the book and the RD Sharma class 12 solutions chapter 18 exercise 18.23 contains 15 questions that acknowledges the basic concepts of this chapter. The essential concepts covered in the RD Sharma class 12th exercise 18.23 are, methods of integration, integration by substitute method, few standard results, evaluation of integrals, special integrals method.

Benefits of using the RD Sharma class 12th exercise 18.23 are given below:-

Any student who has passed the board exams with great scores highly praises the RD Sharma class 12th exercise 18.23 for the expert questions provided in the book helped them the most.
The RD Sharma class 12 chapter 18 exercise 18.23 solution is helpful because it also gives you solved questions, so if you find any difficulty in solving any pattern of question you can refer to these solved questions.
The most interesting thing about these solutions is that it completely corresponds with the syllabus of NCERT, therefore it is important to solve questions from the RD Sharma solutions.
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Frequently Asked Questions (FAQs)

1. Mention a few properties of indefinite integrals?

Few properties of the indefinite integrals are as follows:

∫ 1 dx = x + C
∫ a dx = ax + C
∫ xn dx = ((xn+1)/(n+1)) + C ; n ≠ 1
Integration of trigonometric functions and many more.

2. Why are constants added in indefinite integrals?

We add constants in indefinite integrals in order to include all antiderivatives of f(x).

3. What is the importance of C in indefinite integrals?

The basic importance of C is that it lets the indefinite integrals express the general form of antiderivatives.

4. Is chapter 18 an important chapter for a board exam?

Yes, each and every chapter of the maths subject is important for class 12 students and also it helps you score high

5. From which store can I buy the RD Sharma solutions?

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