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Given is the following function:


With partial integration I found following antiderivative:

$\int\!\sinh(2x)\sin(4x)\mathrm{d}x$ $=\frac{1}{2}\cosh(2x)\sin(4x)-\int\!\frac{1}{2}\cosh(2x)4\cos(4x)$ $=\frac{1}{2}\cosh(2x)\sin(4x)-2\int\!\cosh(2x)\cos(4x)$ $=\frac{1}{2}\cosh(2x)\sin(4x)-2\frac{1}{2}\sinh(2x)\cos(4x)-2\int\!-\frac{1}{2}\sinh(2x)4\sin(4x)$ $=\frac{1}{2}\cosh(2x)\sin(4x)-\sinh(2x)\cos(4x)+4\int\!\sinh(2x)\sin(4x)$ $=-\frac{1}{3}(\frac{1}{2}\cosh(2x)\sin(4x)-\sinh(2x)\cos(4x))$

Nevertheless WolframAlpha has found another solution.

What did I do wrong?

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up vote 1 down vote accepted

I think you missed a sign when differentiating $\cos(4x)$ while doing your second integration by parts. This will flip the sign next to the integral so you have $-4$ instead of $+4$. That will then change the $-1/3$ to $1/5$ and you're done (it matches wolfram).

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