Question

Evaluate $\int {\sin }^{3}x{\cos }^{2}xdx$

Answer 1

${\int }^{}{\sin }^{3}x{\cos }^{2}xdx$

$={\int }^{}{\sin }^{2}x{\cos }^{2}x\sin xdx$

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

putting $\cos x=t$

$\Rightarrow \sin xdx=-dt$

$=-{\int }^{}\left(1-t^2\right)t^{2}dt$

$=-{\int }^{}\left(t^2-t^4\right)dt$

$=-\left[\frac{t^3}{3}-\frac{t^5}{5}\right]+c$

$=\frac{{\cos }^{5}x}{5}-\frac{{\cos }^{3}x}{3}+c$