Question

Integrate:

Answer 1

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

Answer 2

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

Answer 3

maybe IBP?

Answer 4

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

Answer 5

Ok, ill give it a shot

Answer 6

So...

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

\[=\int\limits_{}^{}\frac{ 28\tan^2(\theta)\sec^2(\theta) }{ \sqrt{7(\sec^2(\theta)} }\]

Answer 7

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

Answer 8

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?

Answer 9

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

Answer 10

sines and cosines?

Answer 11

Im trying to do some weird u-sub

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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?

Answer 17

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?\[\large\rm \int\limits \sec^{n}\theta~d \theta=\frac{1}{n-1}\sec^{n-2}\theta \tan \theta+\frac{n-2}{n-1}\int\limits \sec^{n-2}\theta~d \theta\]

Answer 18

nope, new to me

Answer 19

maybe calc III material?

Answer 20

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,
\(\large\rm u=4x\)

Answer 21

hm ok, I see what you did so far there

Answer 22

\[\large\rm u=4x\qquad\qquad\qquad dv=\frac{x}{\sqrt{7+x^2}}dx\]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.

Answer 23

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

Answer 24

Mmmm ok looks good so far

Answer 25

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

Answer 26

cant u sub it can you

Answer 27

I'm not sure.. checking..

Answer 28

No, doesn't seem so :\

Answer 29

Partial Fraction Decomposition doesnt seem to work either

Answer 30

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

Answer 31

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

Answer 32

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.

Answer 33

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.

Answer 34

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

Answer 35

@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."