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Thread: Trigonometric Substitution

  1. #1
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    Exclamation Trigonometric Substitution

    Question :

    $\displaystyle \int x\sqrt{16+x^2}.dx$

    Attempt :
    $\displaystyle \int 4tan\theta .4sec\theta . 4 sec^2\theta$

    How to continue?

    PLease answer in steps , Thank you
    Last edited by mj.alawami; Mar 8th 2010 at 02:38 PM.
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question :

    $\displaystyle \int x\sqrt{16+x^2}$

    Attempt :
    $\displaystyle \int 4tan\theta .4sec\theta . 4 sec^2\theta$

    How to continue?

    PLease answer in steps , Thank you
    1- You should have "dx" in the integral.

    2- $\displaystyle tan(\theta)sec^3(\theta)=\frac{sin(\theta)}{cos^4( \theta)}$. Use $\displaystyle t=cos(\theta)$.

    3- No need for the Trigonometric Substitution. Just use $\displaystyle u=16+x^2$.
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  3. #3
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    Quote Originally Posted by mj.alawami View Post
    Question :

    $\displaystyle \int x\sqrt{16+x^2}$

    Attempt :
    $\displaystyle \int 4tan\theta .4sec\theta . 4 sec^2\theta$

    How to continue?

    PLease answer in steps , Thank you
    Hi mj.alawami,

    here's another way,

    $\displaystyle 16\int{tan\theta\ sec^3\theta}d\theta=\int{u}dv$

    Integrating by parts

    $\displaystyle u=sec^2\theta\ \Rightarrow\ \frac{du}{d\theta}=\frac{du}{dsec\theta}\frac{dsec \theta}{d\theta}=2sec\theta\ sec\theta\ tan\theta=2sec^2\theta\ tan\theta$

    $\displaystyle dv=tan\theta\ sec\theta\ d\theta$

    $\displaystyle v=sec\theta$

    $\displaystyle uv-\int{v}du=sec^3\theta-2\int{sec^3\theta\ tan\theta} d\theta$

    $\displaystyle I=sec^3\theta-2I$

    $\displaystyle 3I=sec^3\theta$

    $\displaystyle I=\frac{1}{3}sec^3\theta$
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