Evaluate $\displaystyle\int \dfrac{\sin x}{\sin 3x} \; dx$
$\begin{aligned}
\text{Let } I & = \int \dfrac{\sin x}{\sin 3x} \; dx \\\\
& = \int \dfrac{\sin x}{3 \sin x - 4 \sin^3 x} \; dx \;\;\; \left[\text{Note: } \sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta\right] \\\\
& = \int \dfrac{\sin x}{3 \sin x \left(1 - \sin^2 x\right) - \sin^3 x} \; dx \\\\
& = \int \dfrac{\sin x}{3 \sin x \cos^2 x - \sin^3 x} \; dx
\end{aligned}$
Now, $\sin x = \dfrac{\sin x / \cos x}{1 / \cos x} = \dfrac{\tan x}{\sec x}$ and $\cos x = \dfrac{1}{\sec x}$
$\begin{aligned}
\therefore \; 3 \sin x \cos^2 x - \sin^3 x & = 3 \times \dfrac{\tan x}{\sec x} \times \dfrac{1}{\sec^2 x} - \dfrac{\tan^3 x}{\sec^3 x} \\\\
& = \dfrac{\tan x \left(3 - \tan^2 x\right)}{\sec^3 x}
\end{aligned}$
$\begin{aligned}
\therefore \; I & = \int \dfrac{\tan x / \sec x}{\tan x \left(3 - \tan^2 x\right) / \sec^3 x} \; dx \\\\
& = \int \dfrac{\sec^2 x}{3 - \tan^2 x} \; dx
\end{aligned}$
Let $\tan x = p$
Then, $\sec^2 x \; dx = dp$
$\begin{aligned}
\therefore \; I & = \int \dfrac{dp}{3 - p^2} \\\\
& = \int \dfrac{dp}{\left(\sqrt{3}\right)^2 - \left(p\right)^2} \\\\
& = \dfrac{1}{2 \sqrt{3}} \log \left|\dfrac{\sqrt{3} + p}{\sqrt{3} - p}\right| + c \;\;\; \left[\text{Note: } \int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a} \log \left|\dfrac{x + a}{x - a}\right| \right] \\\\
& = \dfrac{1}{2 \sqrt{3}} \log \left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c
\end{aligned}$