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Math Help - Evaluate the integral by using substitution prior to integration by parts

  1. #1
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    Exclamation Evaluate the integral by using substitution prior to integration by parts

    ∫ (x^2)/ [square root (x^2 +20)] dx

    this is another problem assigned in my homework and im just really confused on what to do.. on how to use both.. help appreciated!
    Last edited by tennis0216; January 17th 2011 at 06:57 PM.
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  2. #2
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    Quote Originally Posted by tennis0216 View Post
    ∫ (x^2)/ [square root (x^2 +20)] dx

    this is another problem assigned in my homework and im just really confused on what to do.. on how to use both.. help appreciated!
    Trig sub

    x=\sqrt{20}\tan^2(\theta)

    That is the only substitution I can think that will be useful.
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    Quote Originally Posted by dwsmith View Post
    Trig sub

    x=\sqrt{20}\tan^2(\theta)

    That is the only substitution I can think that will be useful.
    i dont even know how substitution would be usefull.. why cant we go directly to integration by parts in this problem?
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  4. #4
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    Quote Originally Posted by tennis0216 View Post
    i dont even know how substitution would be usefull.. why cant we go directly to integration by parts in this problem?
    I am just going by your directions.
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    Quote Originally Posted by dwsmith View Post
    I am just going by your directions.
    no no i am not saying you're wrong.. thank you for the advice by the way! .. im just saying in general.. cant this problem be solved by integration by parts only? why is substitution needed?
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    Quote Originally Posted by tennis0216 View Post
    no no i am not saying you're wrong.. thank you for the advice by the way! .. im just saying in general.. cant this problem be solved by integration by parts only? why is substitution needed?
    You don't need integration by parts. My trig sub will work.
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    Quote Originally Posted by tennis0216 View Post
    no no i am not saying you're wrong.. thank you for the advice by the way! .. im just saying in general.. cant this problem be solved by integration by parts only? why is substitution needed?
    How? What would you use for "u" and what for "dv"?
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    Quote Originally Posted by HallsofIvy View Post
    How? What would you use for "u" and what for "dv"?
    What about \displaystyle u = \frac{1}{2}x and \displaystyle dv = \frac{2x}{x^2 + 20}\,dx?
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    Quote Originally Posted by dwsmith View Post
    Trig sub

    x=\sqrt{20}\tan^2(\theta)

    That is the only substitution I can think that will be useful.
    I think you intended x= \sqrt{20}tan(\theta). That is, the tangent is not squared.

    That way x^2+ 20= 20tan^2(\theta)+ 20= 20sec^2(\theta)
    and dx= \sqrt{20}sec^2(\theta)d\theta

    The integral becomes
    \int \frac{20 tan^2(\theta)}{\sqrt{20}sec(\theta)}(sec^2(theta)d  \theta)
    \sqrt{20}\int tan^2(\theta)sec(\theta)d\theta
    which, switching to sine and cosine, becomes
    \sqrt{20}\int \frac{sin^2(\theta)}{cos^2(\theta)}\frac{1}{cos(\t  heta)}d\theta= \sqrt{20}\int \frac{sin^2(\theta)}{cos^3(\theta)}d\theta

    Since that has an odd power of cosine, I would suggest multiplying both numerator and denominator by cos(\theta), and then letting u= sin(\theta) so that du= cos(\theta)d\theta.

    \sqrt{20}\int \frac{sin^2(\theta)}{cos^4(\theta)}cos(theta(d\the  ta= \sqrt{20}\int \frac{sin^2(\theta)}{(1- sin^2(\theta))^2}cos(\theta)d\theta
    = \sqrt{20}\int \frac{u^2}{(1- u^2)^2}du
    a rational integral that can by done by "partial fractions".
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    Quote Originally Posted by tennis0216 View Post
    ∫ (x^2)/ [square root (x^2 +20)] dx

    this is another problem assigned in my homework and I'm just really confused on what to do.. on how to use both.. help appreciated!
    Another approach to: \displaystyle \int{{x^2}\over{\sqrt{x^2+20}}}\,dx

    I tried the substitution: \displaystyle v={\sqrt{x^2+20} \quad \to \quad dv={{x}\over{\sqrt{x^2+20}}}\,dx\,, which left an x in the integrand, but it did give me an idea for integration by parts.

    \displaystyle \text{Let } dv={{x}\over{\sqrt{x^2+20}}}\,dx \quad \to \quad v={\sqrt{x^2+20} \,,

    \displaystyle \text{and let } u=x\quad \to \quad du=dx\,.

    Then, \displaystyle \int{{x^2}\over{\sqrt{x^2+20}}}\,dx
    \displaystyle = x{\sqrt{x^2+20}-\int{\sqrt{x^2+20}\,dx

    The substitution: \displaystyle x=\sqrt{20}\,\tan(t)\quad\to\quad dx=\sqrt{20}\,\sec^2(t)\,dt\,,

    gives: \displaystyle \int{\sqrt{x^2+20}\,dx=20\int \sec^3(t)\,dt\,.

    This still leaves plenty to do.
    Last edited by SammyS; January 18th 2011 at 11:29 AM.
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  11. #11
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    \displaystyle \int \sec^3(t)\,dt

    Integration by parts:

    \displaystyle \text{Let  }\ u=\sec(t)\quad\to\quad du=\tan(t)\sec(t)\,dt\,,

    \displaystyle \text{and let  }\ dv=\sec^2(t)\,dt\quad\to\quad v=\tan(t)

    \displaystyle \int \sec^3(t)\,dt
    \displaystyle =\tan(t)\sec(t)-\int \tan^2(t)\sec(t)\,dt

    \displaystyle =\tan(t)\sec(t)+\int \sec(t)\,dt-\int \sec^3(t)\,dt

    Therefore,     \displaystyle  2\int \sec^3(t)\,dt=\tan(t)\sec(t)+\int \sec(t)\,dt
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  12. #12
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    This integral demonstrates the simplicity/efficiency of hyperbolic substitutions!

    Let \displaystyle \varphi := \sinh^{-1}{\frac{x}{\sqrt{20}} \Rightarrow \frac{dx}{d\varphi} = \sqrt{20}\cosh{\varphi} \Rightarrow dx = \sqrt{20}\cosh{\varphi}\;{d\varphi} <br />
, then:

    \displaystyle \rhd  \int \frac{x^2}{\sqrt{x^2+20}}\;{dx}   = \int \frac{\left(\sqrt{20}\sinh{\varphi}\right)^2\left(  \sqrt{20}\cosh{\varphi}\right)}{\sqrt{\left(\sqrt{  20}\sinh{\varphi}\right)^2+20}}\;{d\varphi} =  \int \frac{\left({20}\sinh^2{\varphi}\right)\left(\sqrt  {20}\cosh{\varphi}\right)}{\left(\sqrt{20}\cosh{\v  arphi}\right)}\;{d\varphi}
     \displaystyle  = \int \left(10\cosh{2\varphi}-10\right) \;{d\varphi} = 5\sinh{2\varphi}-10{\varphi}+k \displaystyle  = 5\sinh\left(2\sinh^{-1}{\frac{x}{\sqrt{20}}\right)-10\sinh^{-1}{\frac{x}{\sqrt{20}}+k.
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