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

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SUMMARY

The discussion focuses on simplifying the triple integral of cos(u + v + w) over the limits from 0 to π. The user initially attempted u substitution but encountered complexity. By substituting x = u + v + w, the integral simplifies to -sin(u + v + w) evaluated at the limits, yielding 2sin(v + w). The user seeks clarification on the identity sin(x + π) = -sin(x), recognizing it as a shift in the sine function's graph.

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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?
 

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