Evaluate $\int \sin^4 x \; \cos^2 x \; dx$
Let $I = \int \sin^4 x \; \cos^2 x \; dx$ $\;\;\; \cdots$ (1)
Compare equation (1) with $\int \sin^m x \; \cos^n x \; dx$
Here $m = 4$ (even) and $n = 2$ (even)
Now,
$\begin{aligned}
\sin^4 x \; \cos^2 x & = \dfrac{1}{4} \times \left(4 \; \sin^2 x \; \cos^2 x\right) \times \sin^2 x \\\\
& = \dfrac{1}{4} \times \left(2 \; \sin x \; \cos x\right)^2 \times \sin^2 x \\\\
& = \dfrac{1}{4} \times \sin^2 2x \times \sin^2 x \\\\
& = \dfrac{1}{4} \; \left(\dfrac{1 - \cos 4x}{2}\right) \; \left(\dfrac{1 - \cos 2 x}{2}\right) \;\;\; \left[\text{Note: } \sin^2 \theta = \dfrac{1 - \cos 2 \theta}{2} \right] \\\\
& = \dfrac{1}{16} \; \left(1 - \cos 2 x - \cos 4 x + \cos 2x \; \cos 4x\right) \\\\
& = \dfrac{1}{16} \; \left(1 - \cos 2 x - \cos 4 x + \dfrac{2 \cos 2 x \; \cos 4 x}{2}\right) \\\\
& \left[\text{Note: } 2 \cos C \; \cos D = \cos \left(C + D\right) + \cos \left(C - D\right) \right] \\\\
& = \dfrac{1}{32} \; \left(2 - 2 \cos 2 x - 2 \cos 4 x + \cos 6 x + \cos 2 x\right) \\\\
& = \dfrac{1}{32} \; \left(\cos 6 x - 2 \cos 4 x - \cos 2 x + 2\right) \;\;\; \cdots (2)
\end{aligned}$
In view of equation (2), equation (1) becomes
$\begin{aligned}
I & = \dfrac{1}{32} \int \left(\cos 6 x - 2 \cos 4 x - \cos 2 x + 2\right) \; dx \\\\
& = \dfrac{1}{32} \; \left\{\int \cos 6 x \; dx - 2 \int \cos 4 x \; dx - \int \cos 2 x \; dx + 2 \int dx \right\} \\\\
& = \dfrac{1}{32} \; \left\{\dfrac{\sin 6 x}{6} - \dfrac{\sin 4 x}{2} - \dfrac{\sin 2 x}{2} + 2 x \right\} + c
\end{aligned}$