Question

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

Answer 1

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