Answer
$\frac{1}{3}$
Work Step by Step
The area between $y=sinx$ and $y=sin^{3}x$ from $x=0$ to $x=\frac{\pi}{2}$ is determined by doing greater minus lesser:
$A=\int_{0}^{\frac{\pi}{2}}(sinx-sin^{3}x)dx$
$A=\int_{0}^{\frac{\pi}{2}}sinx(1-sin^{2}x)dx$
Through the Pythagorean identity $sin^{2}x+cos^{2}x=1$, $1-sin^{2}x=cos^{2}x$
$A=\int_{0}^{\frac{\pi}{2}}sinx(cos^{2}x)dx$
$let$ $u = cosx$
$du=-sinxdx$
$-du=sinxdx$
$u(0)=1$
$u(\frac{\pi}{2})=0$
$A=-\int_{1}^{0}u^{2}du$
$A=-[\frac{1}{3}u^{3}]_{1}^{0}$
$A=-({\frac{1}{3}}(0)^{3}-\frac{1}{3}(1)^{3})$
$A=-(0-\frac{1}{3})$
$A=\frac{1}{3}$
