Question

Evaluate: $\int \frac{{\sec }^{2}x}{{\tan }^{2}x+4}dx$

Answer 1

$I=\int \frac{{\sec }^{2}x}{{\tan }^{2}x+4}dx$

Let $\tan =t$

${\sec }^{2}xdx=dt$

$I=\int \frac{dt}{t^2+4}$

$=\int \frac{dt}{t^2+(2{)}^2}$

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

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