Many CAS not able to solve integral

nuggetprime · 2007-06-26, 19:11

Hello,
I tried(with wolfram integrator, mupad & derive) to solve this integral:

SIN(x)·(1 - COS(x))·
!!! ANGLES IN DEGREE!!!

None of the three Computer Algebra Systems were able to do this correct(i set angle options). Can anybody help me?

Thanks,
Nuggetprime

alpertron · 2007-06-26, 22:04

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.

2007-06-26, 19:11	#1
nuggetprime Mar 2007 Austria 100101110₂ Posts	Many CAS not able to solve integral Hello, I tried(with wolfram integrator, mupad & derive) to solve this integral: SIN(x)·(1 - COS(x))· !!! ANGLES IN DEGREE!!! None of the three Computer Algebra Systems were able to do this correct(i set angle options). Can anybody help me? Thanks, Nuggetprime

2007-06-26, 22:04	#2
alpertron Aug 2002 Buenos Aires, Argentina 2526₈ 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

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