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    Hi! I know how to do part (i) but I'm struggling with the second part, thanks!
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     \cos 2x =2\cos^2 x-1
    So  x\cos 2x becomes?
    So the integral becomes ?
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    (Original post by B_9710)
     \cos 2x =2\cos^2 x-1
    So  x\cos 2x becomes?
    So the integral becomes ?
    Yeah I got the double angle formula to sub. it in but I don't know where to go from there!
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    so  \displaystyle \int x\cos 2x dx = \int x(2\cos^2 x-1) dx

     \displaystyle = 2\int x\cos^2 x dx - \int x dx

    Just rearrange these integrals to find  \int x \cos^2 x dx
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    (Original post by rpnom)
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    Hi! I know how to do part (i) but I'm struggling with the second part, thanks!
    Following B_9710's post, integrate x cos^2x by parts, and the second integral is trivial.
 
 
 
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