Question

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

Answer 1

$\int \frac{{\sec }^{2}xdx}{{\tan }^{2}x+4}$

Let $t=\tan x\Rightarrow dt={\sec }^{2}xdx$

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

$=\frac{1}{2}{\tan }^{-1}\frac{x}{2}+c$ using $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$

$\therefore \int \frac{{\sec }^{2}xdx}{{\tan }^{2}x+4}=\frac{1}{2}{\tan }^{-1}\frac{x}{2}+c$ where $c$ is the constant of integration.