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February 1, 2015

February 1, 2015

Posted by **Rox** on Thursday, June 14, 2012 at 10:37pm.

find the integral

- calculus -
**Steve**, Friday, June 15, 2012 at 10:42amone 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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