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Math Help - Basic trigonometric integral question

  1. #1
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    Basic trigonometric integral question

    Just wondering about the validity of this answer obtained via using an identity?
    A second opinion would be much appreciated! Thanks.
    Attached Thumbnails Attached Thumbnails Basic trigonometric integral question-sincubexintegral.png  
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  2. #2
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    Re: Basic trigonometric integral question

    Looks good to me.

    - Hollywood
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  3. #3
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    Re: Basic trigonometric integral question

    Quote Originally Posted by DonGorgon View Post
    Just wondering about the validity of this answer obtained via using an identity?
    A second opinion would be much appreciated! Thanks.
    Even though this solution looks fine, it is NOT the most efficient method in this case. I always tell my students to look for obvious substitutions.

    \displaystyle \begin{align*} \int{\sin^3{(x)}\,dx} &= -\int{-\sin{(x)}\sin^2{(x)}\,dx} \\ &= -\int{-\sin{(x)}\left[ 1 - \cos^2{(x)} \right] dx} \end{align*}

    Now let \displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*} and the integral becomes

    \displaystyle \begin{align*} -\int{-\sin{(x)}\left[ 1 - \cos^2{(x)} \right] dx} &= -\int{ 1 - u^2 \, du } \\ &= - \left( u - \frac{u^3}{3} \right)  +C \\ &= -\cos{(x)} + \frac{\cos^3{(x)}}{3} + C \end{align*}
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