Evaluate the following integrals:

Given I $=\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x$

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

$\Rightarrow I=\int \frac{1}{4 \sin ^{2} x+5 \cos ^{2} x} d x=\int \frac{\sec ^{2} x}{4 \tan ^{2} x+5} d x$

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

$\Rightarrow I=\int \frac{d t}{4 t^{2}+5}=\frac{1}{4} \int \frac{d t}{t^{2}+\left(\frac{5}{4}\right)}$

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

$\Rightarrow \frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\left(\frac{5}{4}\right)}=\frac{1}{4} \times \frac{1}{\frac{\sqrt{5}}{2}} \tan ^{-1}\left(\frac{\mathrm{t}}{\frac{\sqrt{5}}{2}}\right)+\mathrm{c}$

$=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 t}{\sqrt{5}}\right)+c$

$=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right)+c$

$\therefore \mathrm{I}=\int \frac{1}{4 \sin ^{2} \mathrm{x}+5 \cos ^{2} \mathrm{x}} \mathrm{dx}=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{2 \tan \mathrm{x}}{\sqrt{5}}\right)+\mathrm{c}$