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Thread: Integration with Trig

  1. #1
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    Integration with Trig

    Hi,

    If you look on the attachment, I cannot understand how to get from the top equation to the second one. I mean, how does 3sin^2(x).cos(x) become just sin^3(x) ???

    Is it something to do with double angle formulae?

    Somebody please explain!!

    Thanks
    Attached Thumbnails Attached Thumbnails Integration with Trig-capture.jpg  
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by CSG18 View Post
    Hi,

    If you look on the attachment, I cannot understand how to get from the top equation to the second one. I mean, how does 3sin^2(x).cos(x) become just sin^3(x) ???

    Is it something to do with double angle formulae?

    Somebody please explain!!

    Thanks
    Are you familiar with substitution rule in integration? You need to know integration to solve this problem

    $\displaystyle \pi \int_0^\frac{\pi}{2} 3sin^{2}x cosx dx $

    Let $\displaystyle u = sinx $

    then $\displaystyle du = cosx dx $

    so, $\displaystyle dx = \frac{du}{cosx}$

    Substitute the values of u and dx in your integral, you get:

    $\displaystyle \pi \int_0^\frac{\pi}{2} 3u^2 cosx \frac{du}{cosx}$

    = $\displaystyle \pi \int_0^\frac{\pi}{2} 3u^2 du$

    = $\displaystyle \pi [\frac{3u^3}{3}]_0^\frac{\pi}{2}$

    =$\displaystyle \pi [u^3]_0^\frac{\pi}{2} $

    Now substitute $\displaystyle u = sin^{2}x$, and you get

    $\displaystyle \pi [sin^{3}x]_0^\frac{\pi}{2} $

    Then Calculate V with the upper and lower limits provided in the question.
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