Evaluate $\displaystyle\int \dfrac{\sin 4x}{\sin x} \; dx$
$\begin{aligned}
\text{Let } I & = \int \dfrac{\sin 4x}{\sin x} \; dx \\\\
& = \int \dfrac{\sin\left[2\left(2x\right)\right]}{\sin x} \; dx \\\\
& = \int \dfrac{2 \sin 2x \cos 2x}{\sin x} \; dx \;\;\; \left[\text{Note: } \sin 2 \theta = 2 \sin \theta \cos \theta\right] \\\\
& = \int \dfrac{4 \sin x \cos x \cos 2x}{\sin x} \; dx \\\\
& = 4 \int \cos x \cos 2x \; dx \\\\
& = 4 \int \cos x \left(1 - 2 \sin^2 x\right) \; dx \;\; \left[\text{Note: } \cos 2x = 1 - 2 \sin^2 x\right] \\\\
& = 4 \int \cos x \; dx - 8 \int \cos x \; \sin^2 x \; dx \;\;\; \cdots (1)
\end{aligned}$
Consider $\int \cos x \; \sin^2 x \; dx$
Let $\sin x = u$ $\;\;\; \cdots$ (2)
Differentiating equation (2) gives $\cos x \; dx = du$
$\begin{aligned}
\therefore \; \int \cos x \; \sin^2 x \; dx & = \int u^2 \; du \\\\
& = \dfrac{u^3}{3} + c_1 \\\\
& = \dfrac{\sin^3 x}{3} + c_1 \;\;\; \cdots (3)
\end{aligned}$
Also, $\int \cos x \; dx = \sin x + c_2$ $\;\;\; \cdots$ (4)
Substituting equations (3) and (4) in equation (1) gives
$I = 4 \sin x - \dfrac{8}{3} \sin^3 x + c$ where $c = c_1 + c_2$