help with an integral please help

edwardraff

Senior member
Feb 20, 2001
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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
Jan 22, 2002
7,020
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(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.
 

Legendary

Diamond Member
Jan 22, 2002
7,020
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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
 

Soccer55

Golden Member
Jul 9, 2000
1,660
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81
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
Jul 9, 2000
1,660
4
81
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 :p

-Tom

EDIT: Got the sign wrong on the substitution
 

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