Can a Shift Simplify the Triple Integral of cos(u+v+w)?

Nah_Roots
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\int \int \int cos(u + v + w)dudvdw (all integrals go from 0 to pi).

I've tried using u substitution for each integral but I end up with a huge integral.
 
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I don't see why. To integrate cos(u+v+w)du, let x= u+ v+ w so dx= du. The integral becomes \int cos(x)dx= -sin(x)+ C= -sin(u+ v+ w)+ C. Evaluating that at 0 and pi gives -sin(pi+ v+ w)+ sin(v+ w). But sin(x+ pi)= -sin(x) so that is just 2 sin(v+w).

Now integrate 2sin(v+w) dv by letting x= v+ w so dx=dv.
 
I don't understand sin(x+ pi)= -sin(x). Is that an identity I am forgetting about?
 
Do you know what the graph of y= sin(x) looks like?
 
Oh, I see. It's a shift, correct?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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