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Thread: Integration by Trig substitution

  1. #1
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    Integration by Trig substitution

    Find the integration of :
    (x^3)/(sqrt(x^2+100))


    I did the following work, but it seems my answer is not matching to the Wolfram|Alpha' answer.
    Integration by Trig substitution-capture.png

    int x^3/(sqrt( x^2 +100)) dx - Wolfram|Alpha
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  2. #2
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    Quote Originally Posted by racewithferrari View Post
    Find the integration of :
    (x^3)/(sqrt(x^2+100))


    I did the following work, but it seems my answer is not matching to the Wolfram|Alpha' answer.
    Click image for larger version. 

Name:	Capture.PNG 
Views:	1116 
Size:	267.0 KB 
ID:	15355

    int x^3/(sqrt( x^2 +100)) dx - Wolfram|Alpha
    it is the same,..just in different form,..try to simplify it
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  3. #3
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    Quote Originally Posted by racewithferrari View Post
    Find the integration of :
    (x^3)/(sqrt(x^2+100))


    I did the following work, but it seems my answer is not matching to the Wolfram|Alpha' answer.
    Click image for larger version. 

Name:	Capture.PNG 
Views:	1116 
Size:	267.0 KB 
ID:	15355

    int x^3/(sqrt( x^2 +100)) dx - Wolfram|Alpha
    Don't use a trig sub, use a hyperbolic sub.

    Let $\displaystyle x = 10\sinh{t}$ so that $\displaystyle dx = 10\cosh{t}\,dt$.


    Then the integral becomes

    $\displaystyle \int{\frac{x^3}{\sqrt{x^2 + 100}}\,dx}$

    $\displaystyle = \int{\frac{1000\sinh^3{t}}{\sqrt{(10\sinh{t})^2 + 100}}\,10\cosh{t}\,dt}$

    $\displaystyle = \int{\frac{1000\sinh^3{t}}{10\cosh{t}}\,10\cosh{t} \,dt}$

    $\displaystyle = \int{1000\sinh^3{t}\,dt}$

    $\displaystyle = \int{1000\sinh{t}\sinh^2{t}\,dt}$

    $\displaystyle = \int{1000\sinh{t}(\cosh^2{t} - 1)\,dt}$

    $\displaystyle = 1000\int{\sinh{t}(\cosh^2{t} - 1)\,dt}$


    Let $\displaystyle u = \cosh{t}$ so that $\displaystyle du = \sinh{t}\,dt$

    The integral becomes

    $\displaystyle 1000\int{u^2 - 1\,du}$

    $\displaystyle = 1000\left(\frac{1}{3}u^3 - u\right) + C$


    Now convert it back to a function of $\displaystyle x$.
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