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How do you find the antiderivative of #(sinx)^2#?

1 Answer

Jul 2, 2016

# = x/2 -1/4 sin 2x + C #

Explanation:

us the identity #cos 2A = 1 - 2 sin^2 A implies sin^2 A = 1/2(1-cos 2A)#

so

#int dx qquad sin^2 x#

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

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

# = x/2 -1/4 sin 2x + C #

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