The Student Room Group
Original post by Ramin Gorji
Basically what it says above. I just cant crack it for some reason, ive tried opening up the bracket which makes it worse. Ive also tried changing the cos2x to 1-2sin^2x and 2cos^2x-1 which still doesnt get me anywhere.

Thanks :smile:

Expanding the brackets is probably the best idea, and recalling that cos4x=2cos22x1\cos 4x = 2\cos ^22x - 1 will be useful.
Reply 2
Well, for simplicity, let u = 2x.

So we're trying to integrate (cosu - 1)^2 = cos^2u - 2cosu + 1

Well, -2cosu + 1 = -2cos(2x) + 1 is trivial to integrate.

What about this cos^2u guy? Well, how do you integrate cos^2x in general? Do the same method but write u instead of x, remembering to convert all the u terms back into x before integrating.
Surely cos2x-1 is 1-2(sinx^2) - 1, which is -2(sinx^2)?
Square that and you'll get 4(sinx^4), goes as follows (minus the 4): http://answers.yahoo.com/question/index?qid=20080115170410AA0eFMZ
Reply 4
Original post by dragonmeister
Surely cos2x-1 is 1-2(sinx^2) - 1, which is -2(sinx^2)?
Square that and you'll get 4(sinx^4), goes as follows (minus the 4): http://answers.yahoo.com/question/index?qid=20080115170410AA0eFMZ


You realise that solution put it back in terms of cos2x\cos 2x again before proceeding? [And then it followed what Farhan/Swayum did.]
Original post by Swayum
Well, for simplicity, let u = 2x.

So we're trying to integrate (cosu - 1)^2 = cos^2u - 2cosu + 1

Well, -2cosu + 1 = -2cos(2x) + 1 is trivial to integrate.

What about this cos^2u guy? Well, how do you integrate cos^2x in general? Do the same method but write u instead of x, remembering to convert all the u terms back into x before integrating.


Can you show your working out please? Because I don't understand what you're saying. :dontknow:
Reply 6
Original post by Scopeowner
Can you show your working out please? Because I don't understand what you're saying. :dontknow:


very optimistic to expect an answer after 4 years ...
Yeah I'm gonna close this thread.

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