Sin[x]sin[2x]x dx again.

To solve the integral [tex]\mbox{$\int x\sin x \sin 2x \,dx$}[/tex], let's first solve [tex]\mbox{$\int \sin x \sin 2x \,dx$.}[/tex]

[tex]
\begin{equation*}
\begin{split}
\int \sin{x}\sin{2x}
&= \int 2 \sin^2 x \cos x\,dx \\
&= \int 2 u^2 du \quad (u = \sin x) \\
&= \frac{2}{3} u^3 + C\\
&= \frac{2}{3} \sin^3 x + C
\end{split}
\end{equation*}
[/tex]

Now, to solve [tex]\mbox{$\int x\sin x \sin 2x \,dx$}[/tex] we may now do integration by parts via

[tex]
\begin{equation*}
\begin{split}
u &= x \\
dv &= \sin x \sin 2x \,dx \\
du &= dx \\
v &= \frac{2}{3} \sin^3 x
\end{split}
\end{equation*}
[/tex]

Which yields:

[tex]
\begin{equation*}
\begin{split}
\int x\sin x \sin 2x \,dx
&= x \frac{2}{3} \sin^3 x - \int \frac{2}{3} \sin^3 x \,dx \\
&= \frac{2}{3} x \sin^3 x - \frac{2}{3} \int \sin x (1 - \cos^2 x)\,dx \\
&= \frac{2}{3} x \sin^3 x - \frac{2}{3} \left( \int \sin x \, dx - \int \sin x \cos^2 x \, dx \right) \\
&= \frac{2}{3} x \sin^3 x - \frac{2}{3}\left( -\cos x + \frac{1}{3} \cos^3 x \right) + C \\
&= \frac{2}{3} x \sin^3 x + \frac{2}{3} \cos x - \frac{2}{9} \cos^3 x + C
\end{split}
\end{equation*}
[/tex]

 

Certainly not a professional TeX coding background. Took me what? Half an hour to get it that way? And even with that I couldn't get a nice table for the IBP.