Evaluate the following integrals:

Given I $=\int \frac{1}{13+3 \cos x+4 \sin x} d x$

We know that $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan \frac{x}{2}}$ and $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan 2 \frac{x}{2}}$

$\Rightarrow \int \frac{1}{13+4 \sin x+3 \cos x} d x=\int \frac{1}{13+4\left(\frac{2 \tan \frac{x}{2}}{1+\tan \frac{2 x}{2}}\right)+3\left(\frac{1-\tan \frac{2 x}{2}}{1+\tan \frac{2 x}{2}}\right)} d x$

$=\int \frac{1+\tan ^{2} \frac{x}{2}}{13+13 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}+3-3 \tan ^{2} \frac{x}{2}} d x$

Replacing $1+\tan ^{2} x / 2$ in numerator by $\sec ^{2} x / 2$ and putting $\tan x / 2=t$ and $\sec ^{2} x / 2 d x=2 d t$,

$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{13+13 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}+3-3 \tan ^{2} \frac{x}{2}} d x=\int \frac{\sec ^{2} \frac{x}{2}}{10 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}+16} d x$

$=\int \frac{2 d t}{10 t^{2}+8 t+16}$

$=\frac{2}{10} \int \frac{1}{t^{2}+\frac{4}{5} t+\frac{8}{5}} d t$

$=\frac{1}{5} \int \frac{1}{\left(t+\frac{2}{5}\right)^{2}+\frac{6^{2}}{5}} d t$

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

$\Rightarrow \frac{1}{5} \int \frac{1}{\left(t+\frac{2}{5}\right)^{2}+\frac{6}{5}^{2}} d t=\frac{1}{5}\left(\frac{1}{\frac{6}{5}}\right) \tan ^{-1} \frac{t+\frac{2}{5}}{\frac{6}{5}}+c$

$=\frac{1}{6} \tan ^{-1}\left(\frac{5 \tan \frac{x}{2}+2}{6}\right)+c$

$\therefore \mathrm{I}=\int \frac{1}{13+3 \cos \mathrm{x}+4 \sin \mathrm{x}} \mathrm{dx}=\frac{1}{6} \tan ^{-1}\left(\frac{5 \tan \frac{\mathrm{x}}{2}+2}{6}\right)+\mathrm{c}$