Question

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

Answer 1

Consider the given integral.

$I=\int \frac{dx}{4{\sin }^{2}x+9{\cos }^{2}x}$

$I=\int \frac{dx}{4{\sin }^{2}x+4{\cos }^{2}x+5{\cos }^{2}x}$

$I=\int \frac{dx}{4+5{\cos }^{2}x}$

$I=\int \frac{dx}{4+\frac{5}{{\sec }^{2}x}}$

$I=\int \frac{{\sec }^{2}xdx}{4{\sec }^{2}x+5}$

$I=\int \frac{{\sec }^{2}xdx}{4(1+{\tan }^{2}x)+5}$

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

$I=\frac{1}{4}\int \frac{{\sec }^{2}xdx}{{\tan }^{2}x+\frac{9}{4}}$

Let $t=\tan x$

$dt={\sec }^{2}xdx$

Therefore,

$I=\frac{1}{4}\int \frac{dt}{t^2+{\left(\frac{3}{2}\right)}^2}$

$I=\frac{1}{4}\int \frac{dt}{{\left(\frac{3}{2}\right)}^2+t^2}$

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

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

$I=\frac{1}{6}{\tan }^{-1}\left(\frac{2t}{3}\right)+C$

On putting the value of $t$, we get

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

Hence, this is the answer.