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Math Help - Help with this arctan integration

  1. #1
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    Help with this arctan integration

    Hello guys. So, I end up facing this integral in a homework, and it looked fine to me. To be honest, I think I have the right answer, but the computer keeps saying it's wrong. I am not sure where's my mistake.

    Help with this arctan integration-integral.png

    The answer I got was 9/4 arctan (x/4) + C. I have absolutely no idea what I did wrong

    Thanks for the help guys!
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  2. #2
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    Re: Help with this arctan integration

    Let x^2 = 16\tan^2\theta. Then x = 4\tan \theta, so dx = 4\sec^2\theta d\theta.

    \int \dfrac{9}{\sqrt{x^2+16}}dx = \int \dfrac{36\sec^2\theta d\theta}{\sqrt{16\sec^2\theta}} = 9\int \sec \theta d\theta

    The integral of \sec \theta is \ln\left|\sec \theta + \tan \theta\right|+C. Since \tan \theta = \dfrac{x}{4} = \dfrac{\text{opp}}{\text{adj}}, you know \sec \theta = \dfrac{\text{hyp}}{\text{adj}} = \dfrac{\sqrt{x^2+16}}{4}. So, the integral is:

    \int \dfrac{9}{\sqrt{x^2+16}}dx = 9\ln\left|\dfrac{\sqrt{x^2+16}}{4} + \dfrac{x}{4}\right|+C = 9\ln\left|\sqrt{x^2+16}+x\right|+C
    Thanks from lguto
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  3. #3
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    Re: Help with this arctan integration

    Thak you so much! Much appreciated!
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