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Math Help - Integration problem ASAP

  1. #1
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    Integration problem ASAP

    Integrate x^2(((x^2)+9)^1/2)
    I have spent an two hours on this problem I'm still no further. I've tried by parts and by substitution.
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  2. #2
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    Re: Integration problem ASAP

    Substitute $\displaystyle \begin{align*} x = 3\tan{ \left( \theta \right) } \implies \mathrm{d}x = 3\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$.
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    Re: Integration problem ASAP

    Sorry I don't understand could you write a full solution - I'd be ever so grateful
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    Re: Integration problem ASAP

    No you can try it yourself. If you get stuck, post exactly what you have done, where you are stuck, and we will help you proceed.
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    Re: Integration problem ASAP

    I don't understand where the 3tan comes from
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    Re: Integration problem ASAP

    If your function has $\displaystyle \begin{align*} a^2 +x^2 \end{align*}$ in it, then a useful substitution is $\displaystyle \begin{align*} x = a\tan{ \left( \theta \right) } \end{align*}$, because it will then simplify to $\displaystyle \begin{align*} a^2 + x^2 = a^2 + \left[ a\tan{ \left( \theta \right) } \right] ^2 = a^2 + a^2 \tan^2{ \left( \theta \right) } = a^2 \left[ 1 + \tan^2{ \left( \theta \right) } \right] = a^2\sec^2{ \left( \theta \right) } \end{align*}$. Why is this useful? Well notice that if $\displaystyle \begin{align*} x = a\tan{ \left( \theta \right) } \end{align*}$ then $\displaystyle \begin{align*} \mathrm{d}x = a\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$. So this extra $\displaystyle \begin{align*} \sec^2{ \left( \theta \right) } \end{align*}$ in the differential will either cancel or combine with the one from the substitution, giving you an EASY function to integrate. Now try it!
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    Re: Integration problem ASAP

    What does a^2 sec^2x equal? I don't understand are we using substitution or what?
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    Re: Integration problem ASAP

    Yes it's substitution, in this case, trigonometric substitution. The problem is that you have written your integral incorrectly. Every integral needs a differential at the end of it. So your integral is actually $\displaystyle \begin{align*} \int{ x^2 \, \sqrt{ x^2 + 9 } \, \mathrm{d}x } \end{align*}$. You need to replace $\displaystyle \begin{align*} x \end{align*}$ with $\displaystyle \begin{align*} 3\tan{ \left( \theta \right) } \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}x \end{align*}$ with $\displaystyle \begin{align*} 3\sec^2{ \left( \theta \right) } \, \mathrm{d} \theta \end{align*}$. When you have done this, try to simplify the integrand.
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    Re: Integration problem ASAP

    When I started my substitution originally I did u=x^2+9 then du=2x dx. Why can't i do it that way?
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    Re: Integration problem ASAP

    Well, I would think you could answer that! When you tried to use that substitution, what went wrong?

    Your original integral was \int x^2(x^2+ 9)^{1/2}dx. If you make the substitution u= x^2+ 9 then du= 2xdx, xdx= du/2. But now, you will need one "x" to use with the differential so that you have \int x (x^2+ 9)^{1/2} (xdx)= \frac{1}{2}\int x u^{1/2}du which still has an "x". You will have to replace that with x= (u- 9)^{1/2} so that once you have converted entirely to the variable u, you have \int (u- 9)^{1/2}u^{1/2} du. That's not any easier to integrate than what you started with.
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    Re: Integration problem ASAP

    If I could answer that I wouldn't have asked the question! Still not understanding any part of this question
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    Re: Integration problem ASAP

    This is one of those problems you can't really get hints for, or possibly even figure out by yourself. As such Prove It has given you the correct substitution to use in post #6. You have a substitution for x in terms of a new variable \theta and what dx is. Using that substitution gives
    \int x^2 \sqrt{x^2 + 9} ~ dx = \int \left ( 3~tan( \theta) \right )^2 \sqrt{ \left ( 3~tan( \theta ) \right ) ^2 + 9}~\left ( 3 ~ sec^2( \theta ) \right )~d \theta

    Now simplify the RHS and see what cancellations/simplifications you can do.

    -Dan
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  13. #13
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    Re: Integration problem ASAP

    Instead if using theta could I use u. I'm always normally use du/dx? Is that the same? Also is (3tan(x))^2 the same as (3sec^2(x))? Could you write (3sec^2(x)) as (3sec(x))^2
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  14. #14
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    Re: Integration problem ASAP

    So far I've got (9tan^2(x))(((9tan^2(x))+9)^1/2)(3sec^2(x)) not sure where to go from here
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    Re: Integration problem ASAP

    Oops instead of the 3 at the end I did 27 at the start
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