MHB How Can You Evaluate a Tricky Integral of Sin^5(3x) dx?

[karush]

$\tiny{8.3.8}$
$\textit{evaluate}$
\begin{align*}\displaystyle
I_{8}&=\int_{0}^{\pi/4}
\sin^5{2x},dx\\
&=-\int_{0}^{\pi/4} (\sin^2(2x))^2 sin(2x) \, dx
\end{align*}
$\textit{set $u=\cos{2x} \therefore du=-2\sin{(2x)} \, dx$
then $u(0)=1$ and $\displaystyle u(\pi/4)=0$}$
\begin{align*}\displaystyle
&=-\int_{1}^{0} (1-u^2)^2 \,du \\
&=-\int_{1}^{0} (u-2u^2+u^4) \,du \\
&=-\left[ \frac{u^2}{2}-\frac{2u^3}{3}+\frac{u^5}{5}\right]_1^{0}\\
&=\frac{4}{15}\textit{ (book answer)}
\end{align*}

$\textit{ok something ? can't get book answer}$

 

[Prove It]

$\tiny{8.3.8}$
$\textit{evaluate}$
\begin{align*}\displaystyle
I_{8}&=\int_{0}^{\pi/4}
\sin^5{2x},dx\\
&=-\int_{0}^{\pi/4} (\sin^2(2x))^2 sin(2x) \, dx
\end{align*}
$\textit{set $u=\cos{2x} \therefore du=-2\sin{(2x)} \, dx$
then $u(0)=1$ and $\displaystyle u(\pi/4)=0$}$
\begin{align*}\displaystyle
&=-\int_{1}^{0} (1-u^2)^2 \,du \\
&=-\int_{1}^{0} (u-2u^2+u^4) \,du \\
&=-\left[ \frac{u^2}{2}-\frac{2u^3}{3}+\frac{u^5}{5}\right]_1^{0}\\
&=\frac{4}{15}\textit{ (book answer)}
\end{align*}

$\textit{ok something ? can't get book answer}$

Remember you said $\displaystyle \begin{align*} \mathrm{d}u = -2\sin{(2\,x)} \,\mathrm{d}x \end{align*}$. You do NOT have that -2 factor in your integrand. What can you do to put it there?

 

[HallsofIvy]

It is also slightly simpler to use the fact that -\int_1^0 f(x)dx= \int_0^1 f(x)dx

 

[karush]

sorry everyone I made some typos in this so I'll just close the post.