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Math Help - Trig Integral

  1. #1
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    Trig Integral

    I've gone over this one but can't see where I am going wrong:

    \displaystyle \int_0^2 x^3\sqrt{x^2 + 4}dx

    Here's my work:

    let x = 2 \tan\theta \rightarrow dx = 2 \sec^2\theta

    \int_0^\frac{\pi}{4} 8 \tan^3\theta2\sec\theta2\sec^2\theta d\theta
    32 \int_0^\frac{\pi}{4} (1-\sec^2\theta)\sec^2\thetasec\thetatan\theta d\theta

    let  u = \sec\theta \rightarrow du = \sec\theta\tan\theta d\theta

    32 \int_1^{\sqrt{2}}  (u^2 - u^4)du

    \frac{32u^3}{3} - \frac{32u^5}{5}\displaystyle\mid_1^\sqrt{2}

    \frac{32 \cdot 2^\frac{3}{2}} {3} - \frac{32 \cdot 2^\frac{5}{2}} {5}-\frac{32}{3} + \frac{32}{5}


    When I plug the original problem into my calculator, I get 10.300645 as an answer. I get 1.767311 on my final answer, which is why I don't think that I am getting to the right place.

    Can anyone help? Thanks.
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  2. #2
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    Re: Trig Integral

    Quote Originally Posted by joatmon View Post
    I've gone over this one but can't see where I am going wrong:

    \displaystyle \int_0^2 x^3\sqrt{x^2 + 4}dx

    Here's my work:

    let x = 2 \tan\theta \rightarrow dx = 2 \sec^2\theta

    \int_0^\frac{\pi}{4} 8 \tan^3\theta2\sec\theta2\sec^2\theta d\theta
    32 \int_0^\frac{\pi}{4} (1-\sec^2\theta)\sec^2\thetasec\thetatan\theta d\theta

    let  u = \sec\theta \rightarrow du = \sec\theta\tan\theta d\theta

    32 \int_1^{\sqrt{2}}  (u^2 - u^4)du

    \frac{32u^3}{3} - \frac{32u^5}{5}\displaystyle\mid_1^\sqrt{2}

    \frac{32 \cdot 2^\frac{3}{2}} {3} - \frac{32 \cdot 2^\frac{5}{2}} {5}-\frac{32}{3} + \frac{32}{5}


    When I plug the original problem into my calculator, I get 10.300645 as an answer. I get 1.767311 on my final answer, which is why I don't think that I am getting to the right place.

    Can anyone help? Thanks.
    How did you get \displaystyle 8\tan^3{\theta}\cdot \sec{\theta} \cdot 2\sec^2{\theta} to \displaystyle 32(1 - \sec^2{\theta})\sec^2{\theta}?

    Surely it would be \displaystyle 32\tan{\theta}(1 - \sec^2{\theta})\sec^3{\theta}...
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  3. #3
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    Re: Trig Integral

    Yes, you're right. I left off a \sec\theta\tan\theta off of that line, which becomes my du after the next substitution. I think that this is a typo and doesn't affect the rest of the calculation. Do you see it otherwise?

    Thanks for your help.
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    Re: Trig Integral

    The mistake is that you use \tan^2\theta =1-\sec^2 \theta (which is incorrect) instead of \tan^2\theta=\sec^2\theta-1.
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  5. #5
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    Re: Trig Integral

    Quote Originally Posted by joatmon View Post
    Yes, you're right. I left off a \sec\theta\tan\theta off of that line, which becomes my du after the next substitution. I think that this is a typo and doesn't affect the rest of the calculation. Do you see it otherwise?

    Thanks for your help.
    Typos make it difficult for the readers to understand what you have done, which makes it harder to give you the guidance that you need.

    Anyway, MATHNEM has now told you where the mistake is. Good luck.
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  6. #6
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    Re: Trig Integral

    I can't believe that I spent this much time looking at this and didn't see that. Thank you!
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