Question

Solve:
$\int {\sin }^{4}x.{\cos }^{2}xdx$

Answer 1

$\int {\sin }^{4}x\cdot {\cos }^{2}xdx=\int \left({\sin }^{2}x\right)({\sin }^{2}x\cdot {\cos }^{2}x)dx$

$=\int \frac{1}{2}(1-\cos 2x){\left(\frac{\sin 2x}{2}\right)}^{2}dx$

$\cos 2x=1-2{\sin }^{2}x$

${\sin }^{2}x=\frac{1-\cos 2x}{2}$

$\sin 2x=2\sin x\cos x$

The integral becomes

$=\int \frac{1}{2}(1-\cos 2x)\cdot {\left(\frac{\sin 2x}{2}\right)}^{2}dx$

$=\frac{1}{8}\left(\int (1-\cos 2x)\cdot \sin \left(2x\right)dx)\right)$

$=\frac{1}{8}\left(\int ({\sin }^{2}2x-\cos 2x\cdot {\sin }^2(2x\left)\right)dx\right)$

$=\frac{1}{8}\int \frac{1}{2}\cdot 2{\sin }^2\left(2x\right)dx-\int \frac{1}{2}\cdot 2\cos 2x{\sin }^{2}2xdx$

$=\frac{1}{8}\left(\int \frac{1}{2}\cdot (1-\cos (4x\left)\right)dx-\int \frac{1}{2}{\sin }^{2}2xd\sin 2x\right)$

$=\frac{1}{16}\left(\int (1-\cos (4x)dx-\int {\sin }^2\left(2x\right)d\sin 2x\right)$

$\frac{d\sin 2x}{dx}=2\cos 2x$

$dsin\left(2x\right)=2\cos 2x\cdot dx$

$=\frac{1}{16}(x-\frac{\sin 4x}{4}-\frac{{\sin }^3\left(2x\right)}{3})+C$.