Question

Integrate the following function: f(x)=$\frac{1}{8{\cos }^{2}x+3{\sin }^{2}x+1}$

Answer 1

$I=\int \frac{1}{8{\cos }^{2}x+3{\sin }^{2}x+1}dx$

$=\int \frac{{\sec }^{2}x}{8+3{\tan }^{2}x+{\sec }^{2}x}dx$

$=\int \frac{{\sec }^{2}x}{8+3{\tan }^{2}x+1+{\tan }^{2}x}dx$

$=\int \frac{{\sec }^{2}x}{9+4{\tan }^{2}x}dx$

Let $2\tan x=t⟹2{\sec }^{2}xdx=dt$

$I=\frac{1}{2}\int \frac{1}{3^2+t^2}dt$

$=\frac{1}{2}\times \frac{1}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+c$

$=\frac{1}{6}{\tan }^{-1}\left(\frac{2\tan x}{3}\right)+c$ (Ans)