#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.22 Question 5 Maths Textbook Solution.

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)

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

$=\int \frac{1}{1+4 t^{2}} d t$

$=\frac{1}{4} \int \frac{1}{\left(\frac{1}{2}\right)^{2}+t^{2}} d t$

$=\frac{1}{4} \times \frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{t}{\frac{1}{2}}\right)+c$                                                                $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$

$=\frac{1}{2} \tan ^{-1}(2 \tan x)+c$