edwardraff

Senior member
what is the indefinite integral of y=cos(x^2)^2

to get rid of any confusion y= cos squared of x squared

I am kind of unsure since integration by parts does not seem to work, and I can't see how to apply the integral of sin(u)^2 from the back of the book to this problem. If anyone has any ideas please help.

Legendary

Diamond Member
(sin(x^2)^3) / 6x
Did you need work? If you need work maybe it's a uv-integral(v du) integral?
I just worked backwards.

edwardraff

Senior member
is that so if anyone could really explain how to do this that would be great

edwardraff

Senior member
work would be GREAT

Legendary

Diamond Member
hmm work...
damn i can't even think of the work...i took calc last year and this is how I did all of the ones I couldn't do work for - just working backwards. Sorry

edwardraff

Senior member
could you show me how you worked backwards

edwardraff

Senior member
it does not seem to work 100% to me

YGPM

Soccer55

Golden Member
Originally posted by: edwardraff
it does not seem to work 100% to me
If you take the derivative of what Legendary posted, it will work out to [cos(x^2)]^2 The trick is finding the substitution and/or integral formula that will get you there.....and as of right now, I can't remember it. Boo for me

-Tom

Soccer55

Golden Member
I just thought of something.....I know that [cos(x)]^2 = (1 + cos(2x))/2, so could you use u = x^2 and apply the substitution [cos(u)]^2 = (1 - cos(2u))/2? I would make sure it works, but I'm too lazy to do it out right now

-Tom

EDIT: Got the sign wrong on the substitution