# Trig Substitution Verification

• Mar 31st 2010, 11:33 AM
kaiser0792
Trig Substitution Verification
I have just worked the following integral and I believe my text has a wrong answer. If anyone would take the time to verify, I would appreciate it.

Initial Problem: $\displaystyle \int\sqrt{4 + 9x^2} dx$

My solution: $\displaystyle \frac{1}{8}[3x\sqrt{4+9x^2} + 4ln|{\sqrt{4+9x^2}+3x}|] + C$

Book solution: $\displaystyle \frac{1}{6}[3x\sqrt{4+9x^2} + 4ln|{\sqrt{4+9x^2}+3x}|] + C$

As you can see, the solutions differ only by the factor's 1/8 and 1/6.
I would appreciate someone verifying for me.
• Mar 31st 2010, 12:15 PM
zzzoak
please find derivative, to prove, what is correct
• Mar 31st 2010, 12:16 PM
If you gave us working it would be easier but Maple gives...

$\displaystyle \frac{1}{2}x\sqrt{4+9x^2} + \frac{2}{3} \textrm{arcsinh} \left(\frac{3x}{2} \right)$

which would seem the book is right...
• Mar 31st 2010, 12:25 PM
kaiser0792
Thanks. That's what I needed to know. My work was rather lengthy and on 2 pages, so I didn't include it. Thanks again.
• Mar 31st 2010, 12:31 PM
kaiser0792
Back to the drawing board!
• Mar 31st 2010, 12:33 PM
Quote:

Originally Posted by kaiser0792
Back to the drawing board!

Just go over your working again, it's not a massive error so you likely just wrote down the wrong number or something... If you can't find the error post up what substitution you used and if I stop raging at my uni work I might have a look at it.
• Mar 31st 2010, 01:15 PM
zzzoak
Z=$\displaystyle \int\sqrt{4+9x^2}dx$=$\displaystyle x\sqrt{4+9x^2}$-$\displaystyle \int{x}d(\sqrt{4+9x^2})$=
=(1)-$\displaystyle \int\frac{9x^2dx}{\sqrt{4+9x^2}}$=(1)-$\displaystyle \int\frac{9x^2+4-4}{\sqrt{4+9x^2}}dx$=
=(1)-Z+4$\displaystyle \int\frac{dx}{\sqrt{4+9x^2}}$
then we get
2Z=$\displaystyle x\sqrt{4+9x^2}$+4$\displaystyle \int\frac{dx}{\sqrt{4+9x^2}}$
the last int is standart.