Evaluate the following integrals:

Given $I=\int \frac{1}{5-4 \sin x} d x$

We know that $\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan \frac{2}{2}}$

$\Rightarrow \int \frac{1}{5-4 \sin x} d x=\int \frac{1}{5-4\left(\frac{2 \tan \frac{x}{2}}{1+\tan \frac{2 x}{2}}\right)} d x$

$=\int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)-4\left(2 \tan \frac{x}{2}\right)} d x$

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

$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)-4\left(2 \tan \frac{x}{2}\right)} d x=\int \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}-8 \tan \frac{x}{2}} d x$

Putting $\tan x / 2=t$ and $\sec ^{2}(x / 2) d x=2 d t$

$\Rightarrow \int \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}-8 \tan \frac{x}{2}} d x=\int \frac{2 d t}{5+5 t^{2}-8 t}$

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

$=\frac{2}{5} \int \frac{1}{\left(t-\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} d t$

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

$\Rightarrow \frac{2}{5} \int \frac{1}{\left(\mathrm{t}-\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} \mathrm{dt}=\frac{2}{5}\left(\frac{1}{\frac{3}{5}}\right) \tan ^{-1}\left(\frac{\mathrm{t}-\frac{4}{5}}{\frac{3}{5}}\right)+\mathrm{c}$

$=\frac{2}{3} \tan ^{-1}\left(\frac{5 \tan \mathrm{x}-4}{3}\right)+\mathrm{c}$

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