$\int\frac{dx}{(1+5\sin^2 x)}$ is equal to

$\frac{1}{\sqrt{6}}\tan^{-1}(\sqrt{6}\tan x) + C$: C is an arbitrary constant

The correct answer is Option (4) → $\frac{1}{\sqrt{6}}\tan^{-1}(\sqrt{6}\tan x) + C$: C is an arbitrary constant

$\displaystyle \int \frac{dx}{1+5\sin^{2}x}$

$=\int \frac{1}{1+5\frac{\tan^{2}x}{1+\tan^{2}x}}\,dx =\int \frac{1}{\frac{1+6\tan^{2}x}{1+\tan^{2}x}}\,dx =\int \frac{\sec^{2}x}{1+6\tan^{2}x}\,dx$

Let $u=\sqrt{6}\tan x\Rightarrow du=\sqrt{6}\sec^{2}x\,dx$.

$\displaystyle \int \frac{\sec^{2}x}{1+6\tan^{2}x}\,dx =\frac{1}{\sqrt{6}}\int \frac{du}{1+u^{2}} =\frac{1}{\sqrt{6}}\tan^{-1}u+C$

$\displaystyle \frac{1}{\sqrt{6}}\tan^{-1}\!\big(\sqrt{6}\tan x\big)+C$