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

1 Answer

Apr 9, 2018

#I=1/4(2x-sin2x)+c#

Explanation:

Here,

#I=intsin^2xdx#

#=int(1-cos2x)/2dx#

#=1/2int(1-cos2x)dx#

#=1/2[x-(sin2x)/2]+c#

#=1/4(2x-sin2x)+c#

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