Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given is the following function:

$f(x):=\sinh(2x)\sin(4x)$

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?

share|improve this question

1 Answer 1

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.