How do you find the integral of #sin^2(3x)dx#?

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Answer 1 · 2016-07-12T22:02:34

#=1/2x - 1/12sin6x + C#

#int \ sin^2 (3x) \ dx#

small book keeping gesture is to make the sub #u = 3x, du = 3 dx#

#1/3int \ sin^2 (u) \ du#

then we use the cosine double angle formulae

#cos 2A = 1 - 2 sin^2 A#

so # sin^2 A = (1 - cos 2A)/2#

#=1/6int \ 1 - cos 2u \ du#

#=1/6( u - 1/2sin 2u ) + C#

#=1/6( 3x - 1/2sin( 2*3x) ) + C#

#=1/2x - 1/12sin6x + C#