Evaluate the integral. $\int sin^3 (5x) dx$ The bounds are from $(\pi/5)$ to $0$. I know we use a pythagorean identity for this. So $u=\sin^2 (x)$ and $du = \sin (2x) dx$. But I'm helping trouble solving this problem. 

\begin{align*}
\int \sin^3(5x) \, dx &= \int \sin^2(5x) \cdot \sin(5x) \, dx \\
&= \int \left( 1-\cos^2(5x) \right)\sin(5x) \, dx \\
&= \int \left( \sin(5x)-\cos^2(5x)\sin(5x) \right) \, dx \\
&= -\frac{\cos(5x)}{5}-\int \cos^2(5x)\sin(5x) \, dx \\
\end{align*}
For the second integral, let $u=\cos(5x)$.
\begin{align*}
 u &=\cos(5x) \\
\Rightarrow  du &=-5\sin(5x) \, dx \\
\Rightarrow  -\frac{1}{5} du &= \sin(5x) \, dx.
\end{align*}
Hence,
\begin{align*}
-\int \cos^2(5x)\sin(5x) \, dx &=\frac{1}{5}\int u^2 \, du \\
&=\frac{u^3}{15}+c \\
&=\frac{\cos^3(5x)}{15}+c.
\end{align*}
Putting it all together,
$$\int \sin^3(5x) \, dx =-\frac{\cos(5x)}{5}+\frac{\cos^3(5x)}{15}+c.$$
I leave you to evaluate your limits.
