What is the antiderivative of #(sin(x))^2#?

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Answer 1 · 2016-08-31T16:33:44

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

use the handy double-angle formula

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

last one gives us #sin^2 A = 1/2(1- cos 2A)#

So

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

# = 1/2 int 1 - cos 2x dx#

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

which you can flip using the other double-angle formula: #sin 2A = 2 sin A cos A#

# = 1/2 (x - sin x cos x) + C#