Let I = \(\int \frac{\sin x}{\sin 3x}dx\)
\(= \int \frac{\sin x}{3\sin x - 4\sin^3x}dx\)
\(= \int \frac{\sin x}{\sin x(3 - 4\sin^2 x)}dx\)
\(= \int \frac 1{3 - 4\sin^2 x}dx\)
Dividing both numerator and denominator by cos2x, we get
\(I = \int \frac {\sec^2x}{3\sec^2 x - 4\tan ^2 x}dx\)
\(=\int \frac{\sec^2 x}{3(1 + \tan ^2 x) - 4\tan ^2 x} dx\)
\(= \int \frac {\sec^2 x}{3 - \tan ^2 x}dx\)
Put tan x = t
∴ sec2x dx = dt
\(I = \int \frac{dt}{3 - t^2}\)
\(= \int \frac{dt}{(\sqrt 3)^2 - t^2}\)
\( = \int \frac{1}{(\sqrt 3)^2 - t^2}dt\)
\(= \frac 1{2\sqrt 3} \log\left|\frac{\sqrt 3 + t}{\sqrt 3 - t}\right|+ c\)
\(= \frac 1{2\sqrt 3} \log\left|\frac{\sqrt 3 + \tan x}{\sqrt 3 - \tan x}\right|+ c\)
