∫dx/(2sin^2x+5cos^2x) is equal to

The correct answer is Option (4) → $\frac{1}{\sqrt{10}}\tan^-1\left(\frac{\sqrt{2}\tan x}{\sqrt{5}}\right)+C$: C is an arbitrary constant

Given $\displaystyle \int \frac{dx}{2\sin^2x+5\cos^2x}$.

Divide numerator and denominator by $\cos^2x$:

$\int \frac{\sec^2x\,dx}{2\tan^2x+5}$.

Let $\tan x=t \Rightarrow \sec^2x\,dx=dt$.

Integral becomes $\int \frac{dt}{2t^2+5}=\frac{1}{\sqrt{10}}\tan^{-1}\!\left(\frac{\sqrt{2}t}{\sqrt{5}}\right)+C$.

Substitute $t=\tan x$:

$\displaystyle \frac{1}{\sqrt{10}}\tan^{-1}\!\left(\frac{\sqrt{2}\tan x}{\sqrt{5}}\right)+C$