Evaluate the following integrals:

Given $I=\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x$

Dividing the numerator and denominator of the given integrand by $\cos ^{2} x$, we get

$\Rightarrow I=\int \frac{1}{4 \cos ^{2} x+9 \sin ^{2} x} d x=\int \frac{\sec ^{2} x}{4+9 \tan ^{2} x} d x$

Putting $\tan x=t$ and $\sec ^{2} x d x=d t$, we get

$\Rightarrow I=\int \frac{d t}{4+9 t^{2}}=\frac{1}{9} \int \frac{d t}{\frac{4}{9}+t^{2}}$

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

$\Rightarrow \frac{1}{9} \int \frac{\mathrm{dt}}{\frac{4}{9}+\mathrm{t}^{2}}=\frac{1}{9} \times \frac{1}{\frac{2}{3}} \tan ^{-1}\left(\frac{\mathrm{t}}{\frac{2}{3}}\right)+\mathrm{c}$

$=\frac{1}{6} \tan ^{-1}\left(\frac{3 t}{2}\right)+c$

$=\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan x}{2}\right)+c$

$\therefore \mathrm{I}=\int \frac{1}{4 \cos ^{2} \mathrm{x}+9 \sin ^{2} \mathrm{x}} \mathrm{dx}=\frac{1}{6} \tan ^{-1}\left(\frac{3 \tan \mathrm{x}}{2}\right)+\mathrm{c}$