RD Sharma Class 12 Exercise 18.22 Indefinite integrals Solutions Maths

RD Sharma Class 12 Exercise 18.22 Indefinite integrals Solutions Maths - Download PDF Free Online

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The Class 12 RD Sharma chapter 18 exercise 18.22 solution is the best solution for the chapter of Indefinite integrals for the students of class 12th. Students of any board can take these solutions for help if they want to score high in the exams. RD Sharma Solutions The RD Sharma class 12th exercise 18.22 is highly recommended for solving the complex chapters like Indefinite integrals, not only teachers but students as well prefer these solutions for thorough practice.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.22
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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.22

Indefinite Integrals Exercise 18.22 Question 1.

Answer: $\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan x}{2}\right)+c$
Given: $\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x$
Hint: Use substitution method
Solution: $\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x$
Dividing numerator and denominator by $\cos ^{2} x$
$\begin{aligned} &=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{4 \cos ^{2} x}{\cos ^{2} x}+\frac{9 \sin ^{2} x}{\cos ^{2} x}} d x \\ &=\int \frac{\sec ^{2} x d x}{4+9 \tan ^{2} x} \end{aligned}$
$Now,Let$
$\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}$ (Differentiating w.r.t to x)
$\text { So, } \int \frac{d t}{4+9 t^{2}}$
$=\frac{1}{9} \int \frac{d t}{\frac{4}{9}+t^{2}}$
$=\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1}\left(\frac{t}{\frac{2}{3}}\right)\right]+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan x}{2}\right)+c$ $[\tan x=t]$

Indefinite Integrals Exercise 18.22 Question 2.

Answer: $\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c$
Given: $\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x$
Hint: Divide the numerator and denominator by $\cos ^{2} x$ and then use substitution method
Solution:
$\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x$
Dividing the numerator and denominator by $\cos ^{2} x$
$\Rightarrow \int \frac{\frac{1}{\cos ^{2} x}}{\frac{4 \sin ^{2} x}{\cos ^{2} x}+5 \frac{\cos ^{2} x}{\cos ^{2} x}} d x$
$=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+5} d x$
Let $\begin{gathered} t=\tan x \\ d t=\sec ^{2} x d x \end{gathered}$ (Differentiating w.r.t to x)
Now, $\int \frac{1}{4 t^{2}+5} d t$
$=\frac{1}{4} \int \frac{1}{t^{2}+\frac{5}{4}} d t$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{4} \times \frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\frac{\sqrt{5}}{2}}\right)+c$
$=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c$

Indefinite Integrals Exercise 18.22 Question 3.

Answer: $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x+1}{\sqrt{3}}\right)+c$
Given: $\int \frac{2}{2+\sin 2 x} d x$
Hint: You must know about substitution method
Solution: $\int \frac{2}{2+\sin 2 x} d x$
$\begin{aligned} &=\int \frac{2}{2+2 \sin x \cos x} d x \\ &=\int \frac{1}{1+\sin x \cos x} d x \end{aligned}$
Dividing the numerator and denominator by $\cos ^{2} x$
$\begin{aligned} &=\int \frac{\sec ^{2} x}{\sec ^{2} x+\tan x} d x \\ &=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+\tan x} d x \end{aligned}$
Let
$\begin{aligned} &\tan x=t \\ &\sec ^{2} d x=d t \end{aligned}$ (Differentiating w.r.t x)
Now, $\int \frac{1}{1+t^{2}+t} d t$
$=\int \frac{1}{t^{2}+2 t \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d t$
$=\int \frac{1}{\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d t$

$=\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$

$\begin{aligned} &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)+c \\ &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x+1}{\sqrt{3}}\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.22 Question 4.

Answer: $\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c$
Given: $\int \frac{\cos x}{\cos 3 x} d x$
Hint: Use the formula of $\cos 3 x$
Solution:
$\int \frac{\cos x}{\cos 3 x} d x$
$=\int \frac{\cos x}{4 \cos ^{3} x-3 \cos x} d x$
$\left(\because \cos 3 x=4 \cos ^{3} x-3 \cos x\right)$
$=\int \frac{1}{4 \cos ^{2} x-3} d x$
Dividing numerator and denominator by $\cos ^{2} x$
$\begin{aligned} &=\int \frac{\sec ^{2} x}{4-3 \sec ^{2} x} d x \\ &=\int \frac{\sec ^{2} x}{4-3\left(1+\tan ^{2} x\right)} d x \end{aligned}$
$\left(\because \sec ^{2} x=1+\tan ^{2} x\right)$
$=\int \frac{\sec ^{2} x}{1-3 \tan ^{2} x} d x$
Let
$\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}$ (Differentiating w.r.t x)
$\begin{aligned} &=\int \frac{1}{1-3 t^{2}} d t \\ &=\frac{1}{3} \int \frac{1}{\left(\frac{1}{\sqrt{3}}\right)^{2}-t^{2}} d t \end{aligned}$
$=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{3}}} \log \left|\frac{\frac{1}{\sqrt{3}}+t}{\frac{1}{\sqrt{3}}-t}\right|+c$ $\left[\int \frac{d t}{a^{2}-t^{2}}=\frac{1}{2 a} \log \left|\frac{a+t}{a-t}\right|+c\right]$
$=\frac{1}{6} \times \sqrt{3} \log \left|\frac{1+\sqrt{3} t}{1-\sqrt{3} t}\right|+c$
$=\frac{1}{2 \sqrt{3}} \log \left|\frac{1+\sqrt{3} \tan x}{1-\sqrt{3} \tan x}\right|+c$

Indefinite Integrals Exercise 18.22 Question 5.

Answer: $\frac{1}{2} \tan ^{-1}(2 \tan x)+c$
Given: $\int \frac{1}{1+3 \sin ^{2} x} d x$
Hint: Use the formula of $\sec ^{2} x$ and then apply substitution method
Solution: $\int \frac{1}{1+3 \sin ^{2} x} d x$
Divide numerator and denominator by $\cos ^{2} x$
$=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\cos ^{2} x}+\frac{3 \sin ^{2} x}{\cos ^{2} x}} d x$
$=\int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x} d x$
$=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+3 \tan ^{2} x} d x$
$\left(\sec ^{2} x=1+\tan ^{2} x\right)$
$=\int \frac{\sec ^{2} x}{1+4 \tan ^{2} x} d x$
$Let$
$\mathrm{t}=\tan x$ (Differentiating w.r.t x)
$d t=\sec ^{2} x d x$
$=\int \frac{1}{1+4 t^{2}} d t$
$=\frac{1}{4} \int \frac{1}{\left(\frac{1}{2}\right)^{2}+t^{2}} d t$
$=\frac{1}{4} \times \frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{t}{\frac{1}{2}}\right)+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{2} \tan ^{-1}(2 \tan x)+c$

Indefinite Integrals Exercise 18.22 Question 6.

Answer: $\frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c$
Given: $\int \frac{1}{3+2 \cos ^{2} x} d x$
Hint: Divide Numerator and Denominator by $\cos ^{2} x$ and then apply substitution method.
Solution: $\int \frac{1}{3+2 \cos ^{2} x} d x$
On dividing Numerator and Denominator by $\cos ^{2} x$
$=\int \frac{\sec ^{2} x}{3 \sec ^{2} x+2}$
$=\int \frac{\sec ^{2} x}{3\left(1+\tan ^{2} x\right)+2} d x$
$\left(\sec ^{2} x=1+\tan ^{2} x\right)$
$=\int \frac{\sec ^{2} x}{5+3 \tan ^{2} x} d x$
$Let$
$t=\tan \: \: x$ (Differentiating w.r.t x)
$d t=\sec ^{2} x d x$
$Now$ , $\int \frac{1}{5+3 t^{2}} d t$
$=\frac{1}{3} \int \frac{1}{\left(\left(\sqrt{\frac{5}{3}}\right)^{2}+t^{2}\right)} d t$
$=\frac{1}{3} \times \frac{\sqrt{3}}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{3}}}\right)+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c$

Indefinite Integrals Exercise 18.22 Question 7.

Answer: $\frac{1}{5} \log \left|\frac{\tan x-2}{2 \tan x+1}\right|+c$
Given: $\int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x$
Hint: Apply substitution method and divide Numerator and Denominator by $\cos ^{2} x$
Solution:
$\int \frac{1}{(\sin x-2 \cos x)(2 \sin x+\cos x)} d x$
$=\int \frac{1}{2 \sin ^{2} x+\sin x \cos x-4 \sin x \cos x-2 \cos ^{2} x} d x$
$=\int \frac{1}{2 \sin ^{2} x-3 \sin x \cos x-2 \cos ^{2} x} d x$
On dividing Numerator and Denominator by $\cos ^{2} x$
$=\int \frac{\sec ^{2} x}{2 \tan ^{2} x-3 \tan x-2} d x$
$Let$
$\tan x=t$ (Differentiate w.r.t x)
$\sec ^{2} d x=d t$
$Now,$ $\int \frac{1}{2 t^{2}-3 t-2} d t$
$=\frac{1}{2} \int \frac{1}{t^{2}-\frac{3}{2} t-1} d t$
$=\frac{1}{2} \int \frac{1}{t^{2}-2 t\left(\frac{3}{4}\right)+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-1} d t$
$=\frac{1}{2} \int \frac{1}{\left(t-\frac{3}{4}\right)^{2}-\left(\frac{5}{4}\right)^{2}} d t$
$=\frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \log \left|\frac{t-\frac{3}{4}-\frac{5}{4}}{t-\frac{3}{4}+\frac{5}{4}}\right|+c$ $\left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right]$
$\begin{aligned} &=\frac{1}{2} \times \frac{2}{5} \log \left|\frac{2t-4}{2 t+1}\right|+c \\ &=\frac{1}{5} \log \left|\frac{2\tan x-4}{2 \tan x+1}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.22 Question 8.

Answer: $\tan ^{-1}\left(\tan ^{2} x\right)+c$
Given: $\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$
Hint: Divide Numerator and denominator by $\cos ^{4} x$
Solution:
$\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$ $(\sin 2 x=2 \sin x \cos x)$
$=\int \frac{2 \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
On dividing Numerator and denominator by $\cos ^{4} x$
$=\int \frac{2 \tan x \cdot \sec ^{2} x d x}{\tan ^{4} x+1}$
$Let$
$\begin{aligned} &\tan ^{2} x=t \\ &2 \tan x \sec ^{2} x d x=d t \end{aligned}$ (Differentiate w.r.t x)
$Now,$ $\int \frac{1}{t^{2}+1} d t$
$=\tan ^{-1} t+c$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\tan ^{-1}\left(\tan ^{2} x\right)+c$

Indefinite Integrals Exercise 18.22 Question 10.

Answer: $\frac{1}{2} \log \left|\frac{\tan x}{\tan x+2}\right|+c$
Given: $\int \frac{1}{\sin ^{2} x+\sin 2 x} d x$
Hint: Use the formula of sin 2x and then apply substitution method
Solution: $\int \frac{1}{\sin ^{2} x+\sin 2 x} d x$
$=\int \frac{1}{\sin ^{2} x+2 \sin x \cos x} d x$ $(\sin 2 x=2 \sin x \cos x)$
On dividing numerator and denominator by $\cos ^{2} x$ , we get
$=\int \frac{\sec ^{2} x}{\tan ^{2} x+2 \tan x} d x$
$Let$
$t=\tan x$
$d t=\sec ^{2} x d x$ (Differentiate w.r.t x)
$Now,$ $\int \frac{1}{t^{2}+2 t} d t$
$\begin{aligned} &=\int \frac{1}{t^{2}+2 t+(1)^{2}-(1)^{2}} d t \\ &=\int \frac{1}{(t+1)^{2}-(1)^{2}} d t \end{aligned}$ $\left[\int \frac{d t}{t^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{t-a}{t+a}\right|+c\right]$
$=\frac{1}{2} \log \left|\frac{t+1-1}{t+1+1}\right|+c$
$\begin{aligned} &=\frac{1}{2} \log \left|\frac{t}{t+2}\right|+c \\ &=\frac{1}{2} \log \left|\frac{\tan x}{\tan x+2}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.22 Question 11.

Answer: $\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+c$
Given: $\int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x$
Hint: Divide numerator and denominator by $\cos ^{2} x$ and then use substitution method
Solution:
$\begin{aligned} &\int \frac{1}{\cos 2 x+3 \sin ^{2} x} d x \\ &\left(\cos 2 x=\cos ^{2} x-\sin ^{2} x\right) \\ &=\int \frac{1}{\cos ^{2} x-\sin ^{2} x+3 \sin ^{2} x} d x \\ &=\int \frac{1}{\cos ^{2} x+2 \sin ^{2} x} d x \end{aligned}$
On dividing numerator and denominator by $\cos ^{2} x$ , we get
$=\int \frac{\sec ^{2} x}{1+2 \tan ^{2} x} d x$
$Let$
$\begin{aligned} &\tan x=t \\ &\sec ^{2} x d x=d t \end{aligned}$ (Differentiate w.r.t x)
$Now,$ $\int \frac{1}{1+2 t^{2}} d t$
$=\frac{1}{2} \int \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}+t^{2}} d t$ $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$
$=\frac{1}{2} \times \frac{\sqrt{2}}{1} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)+c$ $=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+c$

RD Sharma class 12 solution of Indefinite Integrals exercise 18.22 contains 11 questions, covered the following topics that are mentioned below:-

Integration of trigonometric and inverse trigonometric functions
Integration of exponential functions

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