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Math Help - Help on integration by parts

  1. #1
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    Help on integration by parts

    Im solving an the integral of x^2/sqrt(x^2-9) using trig substitution. I have it down to 9*INT sec(x)^3 = secxtanx - INT sec(x)^3 + ln(secx + tanx) so I add the one integral over to the other side and get  10*INT sec(x)^3 = secxtanx + ln(secx + tanx) and then I get INT sec(x)^3 = 1/10[secxtanx + ln(secx+tanx)] but my solutions manual has INT sec(x)^3 = 9/2[secxtanx + ln(secx+tanx)] I'm just confused as to why I can't add the other integral over to get 10. It seems with other integration by parts problems you can but for some reason with the sec(x)^3 they don't add it over?
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  2. #2
    Senior Member polymerase's Avatar
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    \int\frac{x^2}{\sqrt{x^2-9}}\;dx

    Make the substitution x=3\sec{u}, so you end up with \int\frac{x^2}{\sqrt{x^2-9}}\;dx=9\int\sec^3{u}\;du

    9\int\sec^3{u}\;du=9\left(\int\tan^2{u}\sec{u}\;du  +\int\sec{u}\;du\right); use by parts on the first term u=\tan{u} and dv=\sec{u}\tan{u}, you will then end up with

    9\int\sec^3{u}\;du=9\left(\tan{u}\sec{u}+\ln{|\sec  {u}+\tan{u}|}\right)-9\int\sec^3{u}

    18\int\sec^3{u}\;du=9\left(\tan{u}\sec{u}+\ln{|\se  c{u}+\tan{u}|}\right)

    Thus, \int\sec^3{u}\;du=\frac{1}{2}\left(\tan{u}\sec{u}+  \ln{|\sec{u}+\tan{u}|}\right), make all the re-subs and final should be...

    \int\frac{x^2}{\sqrt{x^2-9}}\;dx=9\int\sec^3{u}=\frac{x\sqrt{x^2-9}}{2}+\frac{9}{2}\ln\left|\frac{x+\sqrt{x^2-9}}{3}\right|+C
    Last edited by polymerase; October 18th 2008 at 11:19 AM.
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  3. #3
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    Krizalid's Avatar
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    You misread the question, it's actually \int\frac{x^2}{\sqrt{x^2-9}}.

    Here's my approach:

    \int{\frac{x^{2}}{\sqrt{x^{2}-9}}\,dx}=\int{\frac{x^{2}-9+9}{\sqrt{x^{2}-9}}\,dx}=\underbrace{\int{\sqrt{x^{2}-9}\,dx}}_{\alpha }+9\underbrace{\int{\frac{dx}{\sqrt{x^{2}-9}}}}_{\beta }.

    On \alpha apply partial integration as follows,

    \alpha =\int{(x)'\sqrt{x^{2}-9}\,dx}=x\sqrt{x^{2}-9}-\int{\frac{x^{2}}{\sqrt{x^{2}-9}}\,dx}. On \beta put z=x+\sqrt{x^2-9} and that leads to \beta =\ln \left| x+\sqrt{x^{2}-9} \right|+k_{1}. Finally, put these together and we happily get,

    \begin{aligned}<br />
   \int{\frac{x^{2}}{\sqrt{x^{2}-9}}\,dx}&=x\sqrt{x^{2}-9}-\int{\frac{x^{2}}{\sqrt{x^{2}-9}}\,dx}+9\ln \left| x+\sqrt{x^{2}-9} \right|+k_{2} \\ <br />
  2\int{\frac{x^{2}}{\sqrt{x^{2}-9}}\,dx}&=x\sqrt{x^{2}-9}+9\ln \left| x+\sqrt{x^{2}-9} \right|+k_{2} \\ <br />
  \int{\frac{x^{2}}{\sqrt{x^{2}-9}}\,dx}&=\frac{x\sqrt{x^{2}-9}+9\ln \left| x+\sqrt{x^{2}-9} \right|}{2}+k.<br />
\end{aligned}
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  4. #4
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    polymerase you have it sort of right but instead of 1/2 being outside the answer the book has 9/2. Im still confused as to how they got that because you have a 9 outside of the integral originally but how do they get the 1/2?
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  5. #5
    Senior Member polymerase's Avatar
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    c48316, I have made the answer a bit clearer now, see if you can understand where I got the 1/2.

    To Krizalid, I didn't mis-read the question, I just mis-typed the question (forgot square root)
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  6. #6
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    Oh ok I think I see now. Basically the strategy I should use is solve for \int sec(x)^3dx and then plug it into whatever im using.
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