OpenStudy (legomyego180):

Integrate:

1 year ago
OpenStudy (legomyego180):

\[\int\limits_{}^{}\frac{ 4x^2 }{ \sqrt{7+x^2} }\]

1 year ago
OpenStudy (legomyego180):

looks like trig sub, but the 7 under the radical makes me have doubts.

1 year ago
OpenStudy (legomyego180):

maybe IBP?

1 year ago
OpenStudy (thomas5267):

Use \(x=\sqrt{7}\tan(\theta)\)?

1 year ago
OpenStudy (legomyego180):

Ok, ill give it a shot

1 year ago
OpenStudy (legomyego180):

So... \[\int\limits_{}^{}\frac{ 4x^2 }{ \sqrt{7+x^2} }dx=\int\limits_{}^{}\frac{ 4*7\tan^2(\theta) }{ \sqrt{7+ (\sqrt{7} \tan(\theta))^2}} \sec^2(\theta)=\int\limits_{}^{}\frac{ 28\tan^2(\theta)\sec^2(\theta) }{ \sqrt{7(1+\tan^2(\theta))}}\] \[=\int\limits_{}^{}\frac{ 28\tan^2(\theta)\sec^2(\theta) }{ \sqrt{7(\sec^2(\theta)} }\]

1 year ago
OpenStudy (legomyego180):

So the point of trig sub is to get rid of the radical I think, but I cant see how it's helping me here yet

1 year ago
zepdrix (zepdrix):

Well if you split up the 7 and the secant,\[\large\rm =\frac{28}{\sqrt7}\int\limits \frac{\tan^2\theta \sec^2\theta~d \theta}{\sqrt{\sec^2\theta}}\]you should be able to take the root of that secant squared, ya?

1 year ago
zepdrix (zepdrix):

\[\large\rm =\frac{28}{\sqrt7}\int\limits\limits \frac{\tan^2\theta \sec^2\theta~d \theta}{\sec \theta}\]Cancel some stuff, and then sines and cosines, ya?

1 year ago
OpenStudy (legomyego180):

sines and cosines?

1 year ago
OpenStudy (legomyego180):

Im trying to do some weird u-sub

1 year ago
OpenStudy (legomyego180):

\[\frac{ 28 }{ \sqrt{7} }\int\limits_{}^{}\sec(\theta)(1-\sec^2(\theta)d(\theta)\]

1 year ago
zepdrix (zepdrix):

Ya.. looks like you'll have to do some secant reduction formula.. that's no fun :)

1 year ago
OpenStudy (legomyego180):

er, identities wrong. should be sec^2(theta)-1

1 year ago
zepdrix (zepdrix):

Distribute the secant,\[\large\rm =4\sqrt7\int\limits \sec^3\theta~d \theta-4\sqrt7\int\limits \sec \theta~d \theta\]

1 year ago
OpenStudy (legomyego180):
1 year ago

OpenStudy (legomyego180):

Sounds fishy, we never learned that. Are you sure I'm doing this problem the right way? Maybe by looking at the answer choices it will give us some insight?

1 year ago
zepdrix (zepdrix):

Recall that the integral of secant gives natural log of some business... So it seems like we're heading in the right direction. But yes, you can do it a different way if this feels uncomfortable. Integration by parts would certainly work. Never seen reduction formula for secant?\[\large\rm \int\limits \sec^{n}\theta~d \theta=\frac{1}{n-1}\sec^{n-2}\theta \tan \theta+\frac{n-2}{n-1}\int\limits \sec^{n-2}\theta~d \theta\]

1 year ago
OpenStudy (legomyego180):

nope, new to me

1 year ago
OpenStudy (legomyego180):

maybe calc III material?

1 year ago
zepdrix (zepdrix):

Naw it's calc 2 :D it's whatev though. Let's try Parts maybe? It might be a little confusing knowing how to break it up properly though.\[\large\rm \int\limits \frac{4x^2dx}{\sqrt{7+x^2}}\]Hmm well I know that if I integrate something like this, \(\large\rm \int \frac{x dx}{\sqrt{7+x^2}}\) then it should work out to something nice like \(\large\rm \sqrt{7+x^2}\) I think... So that's probably what we want for our dv, \(\large\rm dv=\frac{x~dx}{\sqrt{7+x^2}}\) and all the left over goes to the u, \(\large\rm u=4x\)

1 year ago
OpenStudy (legomyego180):

hm ok, I see what you did so far there

1 year ago
zepdrix (zepdrix):

\[\large\rm u=4x\qquad\qquad\qquad dv=\frac{x}{\sqrt{7+x^2}}dx\]Then,\[\large\rm du=4dx\qquad\qquad\qquad v=\sqrt{7+x^2}\]If you're unsure about that v, you can do a little side work, maybe a substitution to work it out.

1 year ago
OpenStudy (legomyego180):

\[4x \sqrt{7+x^2}-\int\limits_{}^{}4 \sqrt{7+x^2}\]

1 year ago
zepdrix (zepdrix):

Mmmm ok looks good so far

1 year ago
zepdrix (zepdrix):

And with a trig sub we'll end up at the same place as before. Ah that's no fun lol.

1 year ago
OpenStudy (legomyego180):

cant u sub it can you

1 year ago
zepdrix (zepdrix):

I'm not sure.. checking..

1 year ago
zepdrix (zepdrix):

No, doesn't seem so :\

1 year ago
OpenStudy (legomyego180):

Partial Fraction Decomposition doesnt seem to work either

1 year ago
zepdrix (zepdrix):

Why do all the options have 2x instead of 4x? Hmm

1 year ago
OpenStudy (legomyego180):

Im going to email my professor about it. Ill tell him what you said about secant reduction and see if there is an alternative method we learned Im not remembering, or maybe I skipped the day we discussed secant reduction lol

1 year ago
zepdrix (zepdrix):

It's a little confusing, so that might be a good idea :D Grr this 2x is really bugging me. I made a boo boo somewhere... I can't find it.

1 year ago
OpenStudy (legomyego180):

Ill let you know what he says! Thanks for the help anyway. I got some good practice on IBP and Trig sub from it regardless.

1 year ago
OpenStudy (thomas5267):

Answer is unambiguously horrible. http://m.wolframalpha.com/input/?i=Integrate%5B%284x%5E2%29%2F%287%2Bx%5E2%29%5E%281%2F2%29%2Cx%5D&x=0&y=0

1 year ago
OpenStudy (legomyego180):

@zepdrix in case you were interested, my professor got back to me: I've included a picture of my work I sent to her. "First, you lost sqrt(7) next to sec T dT. Second, in your last line, you'll want to replace tan^2T with sec^2T-1. Then when you distribute that out, you'll have to use IBP to integrate the sec cubed part. It's a really ugly integral, look in the lectures on Powers of Trig. You can find what that integral equals."

1 year ago