# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{1+3 \sin ^{2} x} d x$

Solution:

Given I $=\int \frac{1}{1+3 \sin ^{2} x} d x$

Divide numerator and denominator by $\cos ^{2} x$,

$\Rightarrow I=\int \frac{1}{1+3 \sin ^{2} x} d x=\int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x} d x$

Replacing $\sec ^{2} x$ in denominator by $1+\tan ^{2} x$,

$\Rightarrow \int \frac{\sec ^{2} x}{\sec ^{2} x+3 \tan ^{2} x} d x=\int \frac{\sec ^{2} x}{1+\tan ^{2} x+3 \tan ^{2} x} d x$

$=\int \frac{\sec ^{2} x}{1+4 \tan ^{2} x} d x$

Putting $\tan x=t$ so that $\sec ^{2} x d x=d t$,

$\Rightarrow \int \frac{\sec ^{2} x}{1+4 \tan ^{2} x} d x=\int \frac{d t}{1+4 t^{2}}$

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

We know that $\int \frac{1}{\mathrm{a}^{2}+\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$

$\Rightarrow \frac{1}{4} \int \frac{1}{\frac{1}{4}+\mathrm{t}^{2}} \mathrm{dt}=\frac{1}{4} \times \frac{1}{2} \tan ^{-1}\left(\frac{\mathrm{t}}{2}\right)+\mathrm{c}$

$=\frac{1}{8} \tan ^{-1}\left(\frac{\tan \mathrm{x}}{2}\right)+\mathrm{c}$

$\therefore \mathrm{I}=\int \frac{1}{1+3 \sin ^{2} \mathrm{x}} \mathrm{dx}=\frac{1}{8} \tan ^{-1}\left(\frac{\tan \mathrm{x}}{2}\right)+\mathrm{c}$