calculus
posted by Rox .
∫ sin(x)sin(3x)dx
find the integral

one way:
sin(3x) = sinx cos2x + cosx sin2x
= sinx(2cos^2(x)1) + 2cosx sinx cosx
= cos^2(x) sinx  sinx + 2cos^2(x) sinx
= 3cos^2(x)sinx  sinx
∫ = cos^3(x) + cos(x)
another way:
sinx sin3x = 1/2 cos(2x)  1/2 cos(4x)
∫ = 1/4 sin2x  1/8 sin4x
a little manipulation of that also yields
sin^3(x)  cos(x)
the various expressions are not identical, but differ only by a constant C
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