Indefinite Integration — Maths STD 12 Science — Question

Let $I=\int \frac{1}{3+5 \cos x} d x$ put $\tan \left(\frac{x}{2}\right)=t$
then $d x=\frac{2}{1+t^2} d t$ and $\cos x=\frac{1-t^2}{1+t^2}$
$I=\int \frac{2 \frac{d t}{1+t^2}}{3+5\left(\frac{1-t^2}{1+t^2}\right)}$
$=2 \int \frac{\frac{d t}{1+t^2}}{\frac{3\left(1+t^2\right)+5\left(1-t^2\right)}{1+t^2}}$
$=2 \int \frac{d t}{3+3 t^2+5-5 t^2}$
$=2 \int \frac{d t}{8-2 t^2}$
$=\int \frac{d t}{2^2-t^2}$
$\begin{aligned} & =\frac{1}{2(2)} \log \left|\frac{2+t}{2-t}\right|+c \\ & =\frac{1}{4} \log \left|\frac{2+\tan \left(\frac{x}{2}\right)}{2-\tan \left(\frac{x}{2}\right)}\right|+c\end{aligned}$