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Old 2007-06-26, 22:04   #2
alpertron
 
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Aug 2002
Buenos Aires, Argentina

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Let x be the angle in degrees, so the angle in radians is:

\frac{\pi}{180}x

So you want to integrate:

\sin \left(\frac{\pi}{180}x\right)\left(1-cos(\frac{\pi}{180}x)\right)\,=\, \sin \left(\frac{\pi}{180}x\right)-\sin \left(\frac{\pi}{180}x\right)\,cos(\frac{\pi}{180}x)

using standard sin and cos functions in radian.

Since \sin(2x) = 2\sin x \cos x we get:

\sin \left(\frac{\pi}{180}x\right)-\frac{1}{2}\sin \left(\frac{\pi}{90}x\right)

And its integral is:

-\frac{180}{\pi}\cos \left(\frac{\pi}{180}x\right)+\frac{45}{\pi}\cos \left(\frac{\pi}{90}x\right)

So the result in degrees is (using cosines in degrees):

-\frac{180}{\pi}\cos (x)+\frac{45}{\pi}\cos (2x)

I don't know the solution given by these packages but the difference between their results and the one I gave above should be a constant. Please notice that there are many ways to represent trigonometric identities so it is possible that two very different looking expressions are equal.

Last fiddled with by alpertron on 2007-06-26 at 22:15
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