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. 
 A: $$\int\sin^{2n+1}mx\ dx=\int(1-\cos^2mx)^n\sin mx\ dx$$
Set $\cos mx=u$

Alternatively, $\displaystyle\sin3x=3\sin x-4\sin^3x\iff\sin^3x=\frac{3\sin x-\sin3x}4$
A: \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.
A: Hint : $\sin(3x) = 3\sin x - 4\sin^3x $
$\implies \sin (15x) = 3 \sin (5x) - 4 \sin^3(  5x)$
$\implies \sin^3(  5x) = \dfrac {1}{4} [ 3 \sin (5x) - \sin (15x) ]$
$\implies \int \sin^3(  5x) dx = \dfrac {1}{4} [ 3 \int\sin (5x)dx - \int \sin (15x) dx]$
