Question

Integration help. *question attached below* will give medal

Answer 1

Can I have help on just part (b) of this question? I have already done part (a) and my answer for that was \[\frac{ 1 }{ x \sqrt{x ^{2}-1} }\]

Answer 2

the result of a is affecting the b part

Answer 3

Yes, I'm aware of that but I don't see the connection :S

Answer 4

I'm afraind your result for a part is incorrect

Answer 5

afraid!

Answer 6

Ahhhh mann :(

Answer 7

http://www.wolframalpha.com/input/?i=arctan%28sqrt%7Bx%5E2-1%7D%29

Answer 8

http://www.wolframalpha.com/input/?i=d%2Fdx+%28tan^%28-1%29sqrt%28x^2-1%29%29

Answer 9

yes! it is correct! i redone it!

Answer 10

Oh thank God :D

Answer 11

hehe lol! I almost blow your work hehe

Answer 12

\[
\int_1^{2}\tan^{-1}\sqrt{x^2-1}dx = \left[ x \tan^{-1}\sqrt{x^2-1} \right ]_1^2 -
\int_1^{2}x\frac{1}{x\sqrt{x^2-1}}dx

\]

Answer 13

yes! they want you to use by part like @aum did

Answer 14

\[
u = \tan^{-1}\sqrt{x^2-1} \\
du = \frac{1}{x\sqrt{x^2-1}}dx ~~~~ \text{(From part a)} \\
dv = dx \\
v = x
\]

Answer 15

I guess. Just do it like normal but actually claim part a was involved. Sorry, didnt mean to be to make the problem worse x_x

Answer 16

@aum , how did we get dv=dx?

Answer 17

With by parts, you're allowed to chose dv to be dx and the rest of the integral to be u. It's usually the selection that is made when integrating inverse trig functions.

Answer 18

Oh, alrighty then

Answer 19

And so how do we integrate the question to show that it equals what they gave us? Can you walk me through step by step?

Answer 20

Alright. Well, we are doing this by parts with the selections given above:
\[u = \tan^{-1}\sqrt{x^{2}-1}\]
\[du = \frac{ 1 }{ x{\sqrt{x^{2}-1}} }dx\]
\[dv = dx\]
\[v = x\]So if you recall the by parts formula:
\[uv -\int\limits_{}^{}vdu\]So putting the pieces together, we would have:
\[xtan^{-1}\sqrt{x^{2}-1}-\int\limits_{1}^{2}\frac{ x }{ x \sqrt{x^{2}-1} }dx\]
\[x \tan^{-1} \sqrt{x^{2}-1} - \int\limits_{1}^{2}\frac{ 1 }{ \sqrt{x^{2}-1} }\]Now the remaining integral can be done through trigonometric substitution

Answer 21

Do you have an idea how you might finish that integral by trig sub?

Answer 22

Hmm, I'm looking through my table of integrals but I don't see any that match that can be substituted..

Answer 23

Have you seen trigonometric substitution before?

Answer 24

No, I havent

Answer 25

Ah. Hmm....I wonder how they might have you do this without that. Well, if you dont have any tables that would help you with this integral, mind if I show you how this might be done with a substitution?

Answer 26

Sure, I don't mind :)

Answer 27

Well, when you have a specific form with an integral, you always have this trigonometric substitution option. Now, the exponents of these forms don't matter, but what you're normally looking for are these 3:
\[u^{2} + a^{2}\]
\[\sqrt{u^{2}-a^{2}}\]
\[\sqrt{a^{2}-u^{2}}\]where a is a constant and u is a variable expression. And by variable expression, I mean it could simply be x or it could be 2x^2, or it could even be something like (x+1), but some sort of variable expression. Now, the substitutions that correspond with these forms are:

If \[u^{2} + a^{2}\]let
\[u= a \tan (\theta)\] and \[\sqrt{u^{2} + a^{2}} = a \sec (\theta)\] the 2nd substitution is extra, but it can speed the integral up a bit.

If
\[\sqrt{u^{2}-a^{2}}\]
let
\[u = a \sec (\theta)\]
and\[\sqrt{u^{2}-a^{2}} = a \tan (\theta)\]So very similar to the first one.
If \[\sqrt{a^{2}-u^{2}}\]
let
\[u = a \sin (\theta)\]and
\[\sqrt{a^{2}-u^{2}} = a \cos (\theta)\]Those are the substitutions we would use. Sorry for the many lines, I havent learned how to use th equation editor yet write it all on one line, lol.

Answer 28

So for us, we have the \[\sqrt{u^{2}-a^{2}} \]form. u = x (which seems arbitrary, but it could easily be something else) and a = 1. Therefore our substitutions would be
\[x =u = \sec (\theta)\]
\[\sqrt{x^{2} - 1} = \tan (\theta)\]
Now we always need to get rid of dx when doing anysort of substituion, so I'll differentiate the first substituion to do that, giving me:
\[dx = du = \sec (\theta) \tan (\theta) d \theta\]So that gives me all I need to substitute. So now I get:
\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{x^{2}-1} }dx= \int\limits_{}^{}\frac{ \sec (\theta) \tan ( \theta) d \theta }{ \tan (\theta) } = \int\limits_{}^{}\sec (\theta) d \theta\]

Answer 29

So either by memory or an integration table, this integral gives:
\[\int\limits_{}^{}\sec (\theta) d \theta = \ln \left| \sec (\theta) + \tan (\theta) \right|\]Now the problem is that our integral was in terms of x, so we don't want to keep this solution. We need to look back at our substitutions in the beginning so we can revert back. Thankfully we have direct substitutions going back because we said earlier:
\[x = \sec (\theta)\]\[\sqrt{x^{2} - 1} = \tan (\theta)\]Substituting these back we get:
\[\ln | x + \sqrt{x^{2} -1} | \] Now adding this on to the previous part of our by parts integral, we have a total integration of:
\[x \tan^{-1} \sqrt{x^{2}-1} - \ln | x + \sqrt{x^{2}-1}|\]from 1 to 2. Now we can plug those limits in and get:
\[2\tan^{-1}\sqrt{3} - \ln|2 + \sqrt{3}| - [\tan^{-1}(0)-\ln|1+0|] = \frac{ 2 \pi }{ 3 }- \ln( 2 + \sqrt{3})\]

Answer 30

Sorry for that to be so long, but it comes out okay, lol x_x

Answer 31

Oh no, it's okay :) I'm happy you took the timeout to explain everything clearly and precisely. Let me just process everything! Thank you soooo much

Answer 32

Yeah, sure :) Hope it makes some sense xD