Question

Integrate the following functions.
$\int \frac{se{c}^{2}x}{\sqrt{ta{n}^{2}x+4}}dx.$

Answer 1

$\int \frac{se{c}^{2}x}{\sqrt{ta{n}^{2}x+4}}dx$
Let $tanx=t\Rightarrow se{c}^{2}x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{se{c}^{2}x}$
$\therefore \int \frac{se{c}^{2}x}{\sqrt{ta{n}^{2}x+4}}dx=\int \frac{se{c}^{2}x}{\sqrt{t^2+4}}\frac{dt}{se{c}^{2}x}=\int \frac{dt}{\sqrt{t^2+2^2}}=log|t+\sqrt{t^2+4}|+C[\because \int \frac{dx}{\sqrt{x^2+a^2}}=log|x+\sqrt{x^2+a^2}|]=log|tanx+\sqrt{ta{n}^{2}x+4}|+C(\because t=tanx)$