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Math Help - Trig Substitution Integral

  1. #1
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    Trig Substitution Integral

    \displaystyle \int_0^\frac{\pi}{2} \frac{\cos t dt}{\sqrt{1 + \sin ^2 t}}

    Here's what I have done:

    let x = \sin t \rightarrow dx = \cos t dt , x(\frac{\pi}{2}) = 1, x(0) = 0

    \int_0^1 \frac {dx}{\sqrt{1+x^2}}

    Visualizing a right triangle with the horizontal side = 1 and the vertical side = x,

    \sec\theta = \sqrt{1+x^2}
    \tan\theta = x \rightarrow dx = \sec^2\theta d\theta

    x = 1 \rightarrow \theta = \frac{\pi}{4} \rightarrow \sec\theta = \sqrt{2}

    x = 0 \rightarrow \theta = 0 \rightarrow \sec\theta = 1


    \int_1^{\sqrt{2}} \frac{\sec^2 \theta d\theta}{\sec\theta}

    \int_1^{\sqrt{2}} \sec\theta d\theta

    \ln \mid \sec\theta + \tan\theta \mid_1^\sqrt{2}

    \ln (\sec\sqrt{2} + \tan\sqrt{2}) - \ln(\sec(1) + \tan(1))


    I thought that I had this, but the answer doesn't check out. Can anybody see what I am missing? Thanks again!
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  2. #2
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    Re: Trig Substitution Integral

    Quote Originally Posted by joatmon View Post
    \displaystyle \int_0^\frac{\pi}{2} \frac{\cos t dt}{\sqrt{1 + \sin ^2 t}}

    Here's what I have done:

    let x = \sin t \rightarrow dx = \cos t dt , x(\frac{\pi}{2}) = 1, x(0) = 0

    \int_0^1 \frac {dx}{\sqrt{1+x^2}}

    Visualizing a right triangle with the horizontal side = 1 and the vertical side = x,

    \sec\theta = \sqrt{1+x^2}
    \tan\theta = x \rightarrow dx = \sec^2\theta d\theta

    x = 1 \rightarrow \theta = \frac{\pi}{4} \rightarrow \sec\theta = \sqrt{2}

    x = 0 \rightarrow \theta = 0 \rightarrow \sec\theta = 1


    \int_1^{\sqrt{2}} \frac{\sec^2 \theta d\theta}{\sec\theta}

    \int_1^{\sqrt{2}} \sec\theta d\theta

    \ln \mid \sec\theta + \tan\theta \mid_1^\sqrt{2}

    \ln (\sec\sqrt{2} + \tan\sqrt{2}) - \ln(\sec(1) + \tan(1))


    I thought that I had this, but the answer doesn't check out. Can anybody see what I am missing? Thanks again!
    When you make the substitution \displaystyle x = \tan{\theta} \implies dx = \sec^2{\theta}\,d\theta, you are converting the integrand to a function of \displaystyle \theta, which means your terminals need to be values of \displaystyle \theta, not \displaystyle \sec{\theta}.

    So it should be \displaystyle \int_0^{\frac{\pi}{4}}{\frac{\sec^2{\theta}}{\sqrt  {1 + \tan^2{\theta}}}\,d\theta}.


    But on an entirely different note, I believe hyperbolic substitution will be easier than trigonometric in this case.

    Let \displaystyle x = \sinh{t} \implies dx = \cosh{t}\,dt.
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  3. #3
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    Re: Trig Substitution Integral

    Thanks!
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