∫displaystyledx/(2sin^2x+5cos^2x) is equal to

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

Dividing the numerator and the denominator by \(\cos^2 x\):

    \(\displaystyle \int \frac{dx}{2\sin^2x+5\cos^2x} = \int \frac{\sec^2 x\,dx}{2\tan^2 x + 5}\)

Let \(t = \tan x\), so \(dt = \sec^2 x\,dx\) (substitution):

    \(\displaystyle = \int \frac{dt}{2t^2 + 5} = \frac{1}{2}\int \frac{dt}{t^2 + \left(\frac{\sqrt{5}}{\sqrt{2}}\right)^2}\)

Using the standard integral \(\int \frac{dt}{t^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{t}{a} + C\) with \(a = \frac{\sqrt{5}}{\sqrt{2}}\):

    \(\displaystyle = \frac{1}{2}\cdot\frac{\sqrt{2}}{\sqrt{5}}\tan^{-1}\left(\frac{\sqrt{2}\,t}{\sqrt{5}}\right) + C\)

Since \(\frac{\sqrt{2}}{2\sqrt{5}} = \frac{1}{\sqrt{2}\sqrt{5}} = \frac{1}{\sqrt{10}}\), and t = tan x,

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

NCERT Exemplar:
Integrals