Question

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

Answer 1

`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`