Evaluate the following integrals:
$\int \frac{1}{1+3 \sin ^{2} x} d x$
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 $\int \frac{1}{\mathrm{a}^{2}+\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$
$\Rightarrow \frac{1}{4} \int \frac{1}{\frac{1}{4}+\mathrm{t}^{2}} \mathrm{dt}=\frac{1}{4} \times \frac{1}{2} \tan ^{-1}\left(\frac{\mathrm{t}}{2}\right)+\mathrm{c}$
$=\frac{1}{8} \tan ^{-1}\left(\frac{\tan \mathrm{x}}{2}\right)+\mathrm{c}$
$\therefore \mathrm{I}=\int \frac{1}{1+3 \sin ^{2} \mathrm{x}} \mathrm{dx}=\frac{1}{8} \tan ^{-1}\left(\frac{\tan \mathrm{x}}{2}\right)+\mathrm{c}$
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