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Math Help - Trigonometric Integrals

  1. #1
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    Trigonometric Integrals

    Okay, I am having major problems using the reduction formulas for sine and cosine. I look at the examples in the book, and I can't figure out where some of the numbers come from. I have a problem if anyone can show me how to do it...

    <br />
fsin^4xcos^2x dx<br />

    I've gotten this far:

    fsin^4x(1-sin^2x)dx

    fsin^4x dx - fsin^6x dx

    I am not sure how to proceed from this point. Any help is appreciated!
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  2. #2
    Member Abu-Khalil's Avatar
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    Let I_n=\int\sin^n xdx, then using partial integration I_6=\int(-\cos x)'\sin^5xdx=-\cos x\sin ^5 x + 5\int \sin^4 x\cos ^2 x dx but \int \sin^4 x\cos ^2 x dx=I_6-I_4 so I_6=-\cos x \sin^5 x + 5I_6-5I_4\Rightarrow I_6=\frac{\cos x \sin^5x + 5I_4}{4}.

    But, I_4= \int(\sin^2 x)^2dx=\int\left(\frac{1-\cos 2x}{2}\right)^2dx=\frac{1}{4}\left\{\int dx-\int 2\cos 2xdx+\int\cos^22xdx\right\} and \int 2\cos 2x = \sin 2x +C, \int\cos^22xdx=\int\frac{1+\cos4x}{2}dx=\frac{x}{2  }+\frac{\sin 4x}{8}+K.

    Putting everything back in order, you'll get your primitive.

    You may also, instead of calculating I_4 like I did, re-integrate by parts and you'll have to calculate I_2 which shouldn't be a problem using that \sin^2 x = \frac{1-\cos 2x}{2}
    Last edited by Abu-Khalil; February 5th 2009 at 09:20 PM.
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