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Math Help - Trigonometric substitution

  1. #1
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    Trigonometric substitution

    Evaluate the indefinite integral.


    I'm having trouble how to start. Do I factor out the x to make it ∫√x√16-x dx?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by emurphy View Post
    Evaluate the indefinite integral.


    I'm having trouble how to start. Do I factor out the x to make it ∫√x√16-x dx?
    Complete the square: 16x-x^2=64-64+16x-x^2=64-(x-8)^2

    So you have \int\sqrt{64-(x-8)^2}\,dx

    Now let x=8+8\sin t, dx=8\cos t\,dt, so you get:

    \int\sqrt{64-64\sin^2t}\cdot8\cos t\,dt=64\int\cos^2t\,dt=64\int\frac{\cos(2t)+1}{2}  \,dt

    Integrate and plug back in for t using t=\sin^{-1}\left(\frac{x-8}{8}\right).
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  3. #3
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    I got this to be my answer


    and the integral came out to be sin2t+x+c is this correct?
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  4. #4
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    Quote Originally Posted by emurphy View Post
    I got this to be my answer


    and the integral came out to be sin2t+x+c is this correct?
    You can check your answer by differentiating it. You can also use this website: Wolfram|Alpha
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  5. #5
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    it didnt have it but could you check the if i did the integration for dt=64\int\frac{\cos(2t)+1}{2}\,dt correctly?
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  6. #6
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    Quote Originally Posted by emurphy View Post
    it didnt have it but could you check the if i did the integration for dt=64\int\frac{\cos(2t)+1}{2}\,dt correctly?
    You can check it at the website whose link I gave you.
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