Question

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

• A. $\frac{1}{2}{\cos }^{2}x+c$
• B. $\frac{1}{3}{\cos }^{3}x+c$
• C. $\frac{1}{2}{\sin }^{2}x+c$
• D. $\frac{1}{3}{\sin }^{3}x+c$

Answer 1

The correct option is D $\frac{1}{3}{\sin }^{3}x+c$

$\int {\sin }^{2}x\cos xdx$
Let $u=\sin x$
$\frac{du}{dx}=\cos x\Rightarrow du=\cos xdx$
By replacing $\sin x=u$ and $\cos x=du$, we have
$\int {\sin }^2\cos xdx=\int u^{2}du$
$=\frac{u^3}{3}+c$
$=\frac{1}{3}{\sin }^{3}x+c$