Question

Integrate $\int \frac{{\sec }^{2}x}{\sqrt{{\tan }^{2}x+4}}dx$

Answer 1

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

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

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

We know that $\int \frac{dx}{\sqrt{x^2+a^2}}=\ln (x+\sqrt{x^2+a^2})+c$

$I=\ln (t+\sqrt{t^2+2^2})+c$

$I=\ln (t+\sqrt{t^2+4})+c$

$I=\ln (\tan x+\sqrt{{\tan }^{2}x+4})+c$ where $t=\tan x$