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Math Help - 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
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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

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

    Let u = sinx

    then du = cosx dx

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

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

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

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

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

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

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

    \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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