sine integral

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• November 12th 2008, 08:15 AM
mindblowing
sine integral
integration x sin x dx
• November 12th 2008, 08:30 AM
Chris L T521
Quote:

Originally Posted by mindblowing
integration x sin x dx

You can do this by parts:

$u=x$ and $\,dv=\sin x\,dx$

Thus, $\,du=\,dx$ and $v=-\cos x$

Now I leave it for you to plug it into the integration by parts formula and simplify:

$\int u\,dv=uv-\int v\,du$

Can you take it from here?

--Chris
• November 12th 2008, 12:52 PM
ThePerfectHacker
Quote:

Originally Posted by mindblowing
integration x sin x dx

$\int x \sin x dx = \int x \left( -\cos x \right)' dx = -x\cos x - \int (x)'(-\cos x) dx = -x \cos x + \sin x + k$