Question

Evaluate : ${\displaystyle \int \frac{1}{3+5\cos x}dx}$

Answer 1

Let ${\displaystyle I=\int \frac{1}{3+5\cos x}dx\text{put}\tan \left(\frac{x}{2}\right)=t}$

then ${\displaystyle dx=\frac{2}{1+t^2}dt\text{and}\cos x=\frac{1-t^2}{1+t^2}}$

${\displaystyle I=\int \frac{2\frac{dt}{1+t^2}}{3+5\left(\frac{1-t^2}{1+t^2}\right)}}$

${\displaystyle =2\int \frac{\frac{dt}{1+t^2}}{\frac{3(1+t^2)+5(1-t^2)}{1+t^2}}}$

${\displaystyle =2\int \frac{dt}{3+3t^2+5-5t^2}}$

${\displaystyle =2\int \frac{dt}{8-2t^2}}$

${\displaystyle =\int \frac{dt}{2^2-t^2} }$

 

${\displaystyle =\frac{1}{2\left(2\right)}\log \left|\frac{2+t}{2-t}\right|+c}$

${\displaystyle =\frac{1}{4}\log \left|\frac{2+\tan \left(\frac{x}{2}\right)}{2-\tan \left(\frac{x}{2}\right)}\right|+c}$