Evaluate : \(\displaystyle\int\frac{\sin x}{\sin 3x}dx\)

Answer: \(\dfrac{1}{2\sqrt{3}}\log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c\)

\(\sin 3x = 3\sin x - 4\sin^3 x = \sin x(3 - 4\sin^2 x)\), so \(I = \displaystyle\int\frac{dx}{3 - 4\sin^2 x}\).

Dividing numerator and denominator by \(\cos^2 x\): \(I = \displaystyle\int\frac{\sec^2 x\,dx}{3\sec^2 x - 4\tan^2 x} = \int\frac{\sec^2 x\,dx}{3 + 3\tan^2 x - 4\tan^2 x} = \int\frac{\sec^2 x\,dx}{3 - \tan^2 x}\).

Put \(\tan x = t\), \(\sec^2 x\,dx = dt\): \(I = \displaystyle\int\frac{dt}{(\sqrt{3})^2 - t^2} = \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3} + t}{\sqrt{3} - t}\right| + c\).

\(\therefore I = \dfrac{1}{2\sqrt{3}}\log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c\).