Question

$\int \frac{1}{4{\sin }^{2}x+9{\cos }^{2}x}dx=\frac{1}{k}{\tan }^{-1}\left(\frac{2}{3}\tan x\right).$ Find the value of $k$.

Answer 1

Let $I=\int \frac{1}{4{\sin }^{2}x+9{\cos }^{2}x}dx$
Multiplying numerator and denominator by ${\sec }^{2}x$ we get
$I=\int \frac{{\sec }^{2}x}{4{\tan }^{2}x+9}dx$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
Therefore
$I=\int \frac{1}{4t^2+9}dt=\frac{1}{6}{\tan }^{-1}\left(\frac{2}{3}\tan x\right)$
$\Rightarrow k=6$