# Anyone want to check this solution on WolframAlpha

• Dec 12th 2009, 07:34 PM
Arturo_026
Anyone want to check this solution on WolframAlpha
Well this is part of a larger problem I'm doing. The thing is that I'm checking my answer on WolframAlpha gives me a different solution. I've gone over the integral several times, but I'm unable to find my error (If there's any).

integral [pi*[(2-sinx)^2 - (2-cosx)^2]] from 0 to pi/4 - Wolfram|Alpha

Thanks. (bow)
• Dec 12th 2009, 11:04 PM
Diemo
Quote:

Originally Posted by Arturo_026
Well this is part of a larger problem I'm doing. The thing is that I'm checking my answer on WolframAlpha gives me a different solution. I've gone over the integral several times, but I'm unable to find my error (If there's any).

integral [pi*[(2-sinx)^2 - (2-cosx)^2]] from 0 to pi/4 - Wolfram|Alpha

Thanks. (bow)

Well, I stuck in 45 instead of pi/4 (to compute it in degrees, and wolfram alpha gave me the correct answer, which is indeed $\displaystyle \frac{1}{2}(-9+8\sqrt{2}) \pi$
To get this you expand the integral out:
$\displaystyle \pi \int{(4-4sinx+sin^2x)-(4-4cosx+cos^2(x)}=$$\displaystyle \int{4-4-4sinx+4cosx +sin^2x -cos^x}$$\displaystyle =\int{4(cosx-sinx)-cos(2x)}$
having left out the dxs and the bounds. Integrate, stick in your boundaries and you should get the same as wolfram alpha.
Note:This is also the same as $\displaystyle 4.5+\frac{8}{\sqrt{2}}$
• Dec 13th 2009, 01:20 PM
Arturo_026
I finally caught my mistake. I forgot to calculate for sin(0) and cos(0) because I got used to neglecting those when solving polynomials since almost always they give you zero.

Thanks.