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Math Help - I dont know how he got this...

  1. #1
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    I dont know how he got this...

    1. \int(sin^2(x))^2 cos^2(x) dx
    2. \int(\frac{1-cos(2x)}{2})^2(\frac{1+cos(2x)}{2}dx
    3. \frac{1}{8}\int1-cos(2x)-cos^2(2x)+cos^3(2x)dx
    4. \frac{1}{8}\int1dx-\frac{1}{8}\int cos(2x)dx-\frac{1}{8}\int cos^2(2x)dx+\frac{1}{8}\int cos^3(2x)dx
    5. \frac{1}{16}x-\frac{1}{64}sin(4x)-\frac{1}{48}sin^3(2x)+C


    What I dont understand is how he got from 2 to 3 and from 4 to 5. For three, I had this:
    \frac{1}{8}\int-c^3-2c^2+c+1dx Where c is a place holder for cos 2x.
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  2. #2
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    Re: I dont know how he got this...

    Check your calculation from step 2 to step 3, since by my reckoning

    (1-\cos(2x))^2(1+\cos(2x))=1-\cos(2x)-\cos^2(2x)+\cos^3(2x)

    For the final step Wikipedia is certainly your friend:

    List of trigonometric identities - Wikipedia, the free encyclopedia

    You can use the half angle, double angle and triple angle identities to make the last two integrations in step 4 manageable.
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  3. #3
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    Re: I dont know how he got this...

    Wow, I wrote down the problem wrong, I had sine and cosine switched. I copied the above straight from the book, so it's correct. I wrote it wrong in my note book. I swear, I need to postpone calculus until I get down basic reading and writing...
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