∫displaystyledx/((1+5sin^2 x)) is equal to

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

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

    \( \displaystyle \int\frac{dx}{1 + 5\sin^2 x} = \int\frac{\sec^2 x\,dx}{\sec^2 x + 5\tan^2 x} \)

Since \(\sec^2 x = 1 + \tan^2 x\):

    \( \displaystyle = \int\frac{\sec^2 x\,dx}{1 + 6\tan^2 x} \)

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

    \( \displaystyle = \int\frac{dt}{1 + (\sqrt{6}\,t)^2} \)

Using the standard integral \(\int\frac{du}{1 + u^2} = \tan^{-1}u\) with \(u = \sqrt{6}\,t\), \(du = \sqrt{6}\,dt\):

    \( \displaystyle = \frac{1}{\sqrt{6}}\tan^{-1}(\sqrt{6}\,t) + C \)

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