Question

How do i take the antiderivative of (sinxcosx)^2?

Answer 1

need any direction.. Maybe i missed this but im completely lost

Answer 2

I went to Wolfram and clicked Show Steps.

http://www.wolframalpha.com/input/?i=integral%20(sinxcosx)%5E2%20dx&t=crmtb01

I hope that can help, I'm not really sure how to give you a good answer on this one, since it looks straight forward but apparently it isn't.

Answer 3

First (sinx*cosx)^2 = sin^2 x*cos^2 x

Then you can use the half angle formula for both sin^2 x and cos^2 x
[sin^2 x= (1-cos2x)/2 && cos^2 x=(1+cos2x)/2]
So,
sin^2 x * cos^2 x dx = (1-cos2x)/2 *(1+cos2x)/2 --->[(a+b)(a-b)= a^2-b^2]
'' '''''''''' =(1-cos^2 2x)/4
Now you can break this apart as (1/4 - cos^2 2x/4) dx
The antiderivative of 1/4 is just x/4
And next we find the antiderivative of the cos^2 2x / 4 by again using the half angle identity that we used before.
So this would be cos^2 2x / 4 = (1+cos 2 2x)/(2*4)
=1+cos4x/8
Next, we have to find the antiderivative of (1+cos4x)/8
(1+cos4x)/8 dx = x/8+sin4x/32 ---> This is just by using the general antiderivative
Therefore, the final answer would be =>x/4 - x/8 - sin4x/32 + C

I hope this helps you.