How do you find the antiderivative of #sin^4(x)#?

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Answer 1 · 2015-04-29T18:51:59

You could use power reduction formulas (double angle)

Because #cos 2 theta = 1-2sin^2 theta = 2cos theta -1#, we get

#sin^2theta = 1/2 (1-cos2 theta)#
and
#cos^2theta = 1/2 (1+cos2 theta)#

#int sin^4x dx = int (sin^2x)^2 dx#

#color(white)"sssssssssss"# # = int [1/2(1-cos2x)]^2 dx#

#color(white)"sssssssssss"# # = 1/4 int (1-2cos2x+ cos^2 2x) dx#

#color(white)"sssssssssss"# # = 1/4 int (1-2cos2x+ 1/2(1+cos4x)) dx#

#color(white)"sssssssssss"# # = 1/4 int (3/2-2cos2x+ 1/2 cos4x) dx#

#color(white)"sssssssssss"# # = 1/4 (3/2 x - sin2x+ 1/8 sin4x) +C#

#color(white)"sssssssssss"# # = 3/8 x- 1/4 sin2x+ 1/32 sin4x +C#