Integrate expression with 3 multiplications $\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}dx$ Specifically I want to integrate the following:
$$\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)}$$
I do know that
$$\frac4{\pi}\int_0^{\frac\pi4}{x\cos(x)\sin(2nx)} = \dfrac{-16\cos(n\pi)}{(4n^2-1)^2{\pi}}$$
I just have no idea how to integrate it. I tried using trig identities to simplify the expression something by parts could be used for, but I got a very wrong answer.

Using $$\cos x \sin 2 n x=\frac{1}{2}[\sin (2 n+1) x+\sin (2 n-1) x],$$
we have
$$
\begin{aligned}
&\quad \int_0^{\frac{\pi}{2}} x \cos x \sin 2 n x d x\\& =\frac{1}{2}\left[\int_0^{\frac{\pi}{2}} x[\sin (2 n+1)+\sin (2 n-1) x] d x\right] \\
& =-\frac{1}{2} \int_0^{\frac{\pi}{2}} x d\left(\frac{\cos (2 n+1) x}{2 n+1}+\frac{\cos (2 n-1) x}{2 n-1}\right) \\
& =-\frac{1}{2}\left(x\left[\frac{\cos (2 n+1) x}{2 n+1}+\frac{\cos (2 n-1) x}{2 n-1}\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{\cos (2 n+1) x}{2 n+1} d x-\frac{\cos (2 n-1) x}{2 n-1} d x\right) \\
& =\frac{1}{2}\left[\frac{\sin (2 n+1) x}{(2 n+1)^2}+\frac{\sin (2 n-1) x}{(2 n-1)^2}\right]_0^{\frac{\pi}{2}}\\&= \frac{(-1)^n}{2\left(4n^2-1\right)^2}\left[(2 n-1)^2-(2 n+1)^2\right]\\&=-\frac{4 n \cos n \pi}{\left(4 n^2-1\right)^2}
\end{aligned}
$$
