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Math Help - integration substitution

  1. #1
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    integration substitution

    Hi , after a substitution with x=3 cos u the answer in the text gives the following...see image

    I can't see how it tidies up to u cos u
    thank you
    Attached Thumbnails Attached Thumbnails integration substitution-jock.jpg  
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  2. #2
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    Re: integration substitution

    Could you please post all of the original question?
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  3. #3
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    Re: integration substitution

    the question is.. by considering the substitution x=3cos u ,evaluateintegration substitution-untitled.jpg
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    Re: integration substitution

    Quote Originally Posted by minicooper58 View Post
    the question is.. by considering the substitution x=3cos u ,evaluateClick image for larger version. 

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    Let \displaystyle \begin{align*} x = 3\cos{u} \implies dx = -3\sin{u}\,du \end{align*}, and noting that when \displaystyle \begin{align*} x = 0, u = \frac{\pi}{2} \end{align*} and when \displaystyle \begin{align*} x = \frac{3}{2}, u = \frac{\pi}{3} \end{align*}, the integral becomes

    \displaystyle \begin{align*} \int_0^{\frac{3}{2}}{\frac{x\arccos{\left(\frac{x}  {3}\right)}}{\sqrt{9 - x^2}}\,dx} &= \int_{\frac{\pi}{2}}^{\frac{\pi}{3}}{\frac{3\cos{u  }\arccos{\left(\frac{3\cos{u}}{3}\right)}}{\sqrt{9 - \left(3\cos{u}\right)^2}}\cdot -3\sin{u}\,du} \\ &= -9\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}{\frac{u\cos{  u}\sin{u}}{\sqrt{9 - 9\cos^2{u}}}\,du} \\ &= 9\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{u\cos{  u}\sin{u}}{\sqrt{9\sin^2{u}}}\,du} \\ &= 9\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{u\cos{  u}\sin{u}}{3\sin{u}}\,du} \\ &= 3\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{u\cos{u}\,du  } \end{align*}

    which you can now solve using Integration by Parts
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  5. #5
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    Re: integration substitution

    The terms on the top are multiplied together so 3sinu cancels out.
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  6. #6
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    Re: integration substitution

    Quote Originally Posted by minicooper58 View Post
    the question is.. by considering the substitution x=3cos u ,evaluateClick image for larger version. 

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    After substitution integral becomes :

    I=-3\cdot \int\limits_{\frac{\pi}{2}}^{\frac{\pi}{3}} u \cdot \cos u\,du=3\cdot \int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} u \cdot \cos u\,du

    which can be solved using integration by parts...
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