Find the area between $\displaystyle \[f(x)=sin^5(x)\]$ and $\displaystyle \[g(x)=sin^3(x)\]$ both with domain $\displaystyle \[[0,\pi]\]$.

On $\displaystyle \[[0,\pi]\]$ the point of intersection is 0 and $\displaystyle \[\frac{\pi}{2}\]$. So,

$\displaystyle \[\int_{0}^{\frac{\pi}{2}}[sin^3(x)-sin^5(x)]dx$

$\displaystyle =\int_{0}^{\frac{\pi}{2}}[(sin^2x)sin(x)-(sin^2x)^2sin(x)]\ dx$

$\displaystyle =\int_{0}^{\frac{\pi}{2}}[(1-cos^2x)sin(x)-(1-cos^2x)^2sin(x)]\ dx$

$\displaystyle u=cos(x); \ du=-sin(x)$

$\displaystyle \int_{0}^{1}[(1-u^2)-(1-u^2)^2]\ du$

$\displaystyle = u - \frac{1}{3}u^3 - u + \frac{2}{3}u^3-\frac{1}{5}u^5\biggr|_{0}^{1}$

$\displaystyle = 1 - \frac{1}{3} - 1 +\frac{2}{3} - \frac{1}{5} = \frac{2}{15}$

However, the answer in the back of the book says the answer is $\displaystyle \frac{4}{15}$. Did I do anything wrong?