#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 6 Maths Textbook Solution.

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)

$d t=\sec ^{2} x d x$

$Now$,$\int \frac{1}{5+3 t^{2}} d t$

$=\frac{1}{3} \int \frac{1}{\left(\left(\sqrt{\frac{5}{3}}\right)^{2}+t^{2}\right)} d t$

$=\frac{1}{3} \times \frac{\sqrt{3}}{\sqrt{5}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{3}}}\right)+c$                                                    $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$

$=\frac{1}{\sqrt{15}} \tan ^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{5}}\right)+c$