Question

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

Answer 1

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

This can be written as

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

$\Rightarrow \int \frac{1}{4{\sin }^{2}x+9{\cos }^{2}x}dx$

This can be written as

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

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

$\Rightarrow \int \frac{dt}{4t^2+9}$

$\Rightarrow \frac{1}{4}\int \frac{dt}{t^2+\frac{9}{4}}$

$\Rightarrow \frac{1}{4}\int \frac{dt}{t^2+(\frac{3}{2}{)}^2}$

$\Rightarrow \frac{1}{4}\times \frac{2}{3}{\tan }^{-1}\left(\frac{2t}{3}\right)+C$

$\Rightarrow \frac{1}{6}{\tan }^{-1}\left(\frac{2\tan x}{3}\right)+C$