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 2007-06-26, 22:04 #2 alpertron     Aug 2002 Buenos Aires, Argentina 22·3·113 Posts 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