# Thread: Help on integration by parts

1. ## Help on integration by parts

Im solving an the integral of $\displaystyle x^2/sqrt(x^2-9)$ using trig substitution. I have it down to $\displaystyle 9*INT sec(x)^3 = secxtanx - INT sec(x)^3 + ln(secx + tanx)$ so I add the one integral over to the other side and get $\displaystyle 10*INT sec(x)^3 = secxtanx + ln(secx + tanx)$ and then I get $\displaystyle INT sec(x)^3 = 1/10[secxtanx + ln(secx+tanx)]$ but my solutions manual has $\displaystyle INT sec(x)^3 = 9/2[secxtanx + ln(secx+tanx)]$ 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?

2. $\displaystyle \int\frac{x^2}{\sqrt{x^2-9}}\;dx$

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

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

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

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

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

$\displaystyle \int\frac{x^2}{\sqrt{x^2-9}}\;dx=9\int\sec^3{u}=\frac{x\sqrt{x^2-9}}{2}+\frac{9}{2}\ln\left|\frac{x+\sqrt{x^2-9}}{3}\right|+C$

3. You misread the question, it's actually $\displaystyle \int\frac{x^2}{\sqrt{x^2-9}}.$

Here's my approach:

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

On $\displaystyle \alpha$ apply partial integration as follows,

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

\displaystyle \begin{aligned} \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} \\ 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} \\ \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. \end{aligned}

4. 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?

5. 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)

6. Oh ok I think I see now. Basically the strategy I should use is solve for $\displaystyle \int sec(x)^3dx$ and then plug it into whatever im using.