3
$\begingroup$

I'm currently studying for my calculus exam, and i've run into a problem that has given me tons of issues. It's not one i've worked on before(or even seen before) so i'm worried if I see one on my exam, I won't know how to solve it properly. I've spent a few hours on it trying different things with no luck whatsoever. The problem is as follows:

$$\int\frac{\sqrt{9x^2-25}}{x}dx$$

I identify this(perhaps incorrectly?) as a secant problem, where: $$3x=5\sec\theta$$ and $$dx=\frac{5}{3}\sec\theta\tan\theta\; d\theta$$

Then, subbing back in I get:

$$\int\frac{5\tan\theta\;\frac{5}{3}\sec\theta\tan\theta}{\frac{5}{3}\sec\theta}\;d\theta$$

Reducing I get(again, maybe incorrectly? My trig isn't that great but i'm working hard at it):

$$\int5\tan^2\theta\;d\theta$$

And finally, using $\tan^2\theta = \sec^2\theta-1$ I get:

$$5\int(\sec^2\theta-1)\;d\theta= 5\tan\theta-5\theta + C$$

Now, subbing back in for theta, where $\tan\theta = \frac{\sqrt{9x^2\;-\;25}}{5}$ , I get:

$$ \sqrt{9x^2-25} - 5arcsec(x) + C$$

However, this is not the answer my professor has listed in the answer key. He gives the solution as:

$$-\frac{3\sqrt{9x^2-25}}{x^2}+\frac{27}{50}arcsec(x) + C$$

Now, his answers have been wrong before but I think I just made a major error along the way. The arcsecant at the end tells me I was kind of on the right track, but the fraction in front of it tells me I messed up big somewhere. Did I make a mistake converting the radical for trig substitution, make a reducing error, or both? Should I have tackled this problem using a different technique? He won't answer my emails, so i'm hoping you fine math wizards can help me figure out what I did wrong.

Sorry for the formatting ahead of time, I wanted to highlight my answers but I still don't know how(the basic mathjax tutorial doesn't seem to include it?). Anyways, thanks ahead of time for any help. I really want to know how to solve these, it's killing me!

$\endgroup$
  • $\begingroup$ Your version looks more correct, except that it is not $5\text{arcsec}(x)$. Closer is $5\text{arcsec}(3x/5)$ except there may be an issue when $x$ is negative. $\endgroup$ – André Nicolas Jun 14 '16 at 18:59
  • $\begingroup$ That is true, good point. There is one mistake for sure. I just can't see how my professor's answer is correct, it doesn't even seem like i'm close... $\endgroup$ – FuegoJohnson Jun 14 '16 at 19:08
  • $\begingroup$ After making the adjustment mentioned in that first comment, have you tried checking your solution (take a derivative) and your professors? $\endgroup$ – user328442 Jun 14 '16 at 19:12
  • $\begingroup$ @FuegoJohnson But everyone makes mistakes. Messy substitutions and all that, when learning calculus the first time, we all make mistakes. Take it from a pro (e.g. not me) ;D $\endgroup$ – Simply Beautiful Art Jun 14 '16 at 19:17
  • $\begingroup$ The good thing is that one can easily check answers by differentiating, which is usually easier. And a note: If one let $u=\sqrt{9x^2-25}$ then the natural form of the answer becomes $$\sqrt{9x^2-25}-5\arctan\bigl(\frac15\sqrt{9x^2-25}\bigr)+C.$$ This just shows that answers might look different. I don't think that is the case for the teacher here, though... $\endgroup$ – mickep Jun 14 '16 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.