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Integration problem

Mathematics
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|dw:1349015981615:dw|
This looks like one of the trig substitutions.
or maybe integration by parts would work.

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Other answers:

|dw:1349016003653:dw|
so I will rewrite it as|dw:1349016056313:dw|
Am I doing it correct so far?
I think it's best if we leave it inside. Something tells me that will have to stay as u^(3/2)
then I don't see with trig sub I am suppose to use
oh nvm
Nope . Use Inverse sec :) or Sine? I forgot what to use but simple as that :D
It is trig, u-sub won't work. Yeah keep it out, it's inverse sin of 6x i believe
|dw:1349016200316:dw|
Its nerp it is sin inverse x/6 +C :)
|dw:1349016274224:dw| is what my solution is suppose to be. Of course I am more interested in how to get to the solution
I think we should use sec instead of sin?! Let x = 6sec (theta)
36Sec^2(theta) - 36 = 36 tan^2(theta)
so the bottom is: sqrt (36 tan^2(theta)) = 6tan^2(theta), yea callisto's right
and top is: dx = 6Sec(theta)Tan(theta)dtheta
so is my given solution incorrect?
yea precal
I have some integrals at the backof my book "Integral involving x^2-a^2; a>0"
\[\int\limits \frac{ 6secutanudu }{ (36\sec^2u - 36)^{3/2} } = \int\limits \frac{ 6secu tanu du }{ 6^3 \tan^3u }\] \[1/36 \int\limits secdu/\tan^2u \] \[1/36 \int\limits \frac{ cosudu }{ \sin^2u? }\]
cos u/ sin^2 u = cscu cotu \[\int \csc u \cot udu = -cscu +C\] Time to draw a triangle...
time to draw a triangle? how is that going to help....I wonder if we are taking the correct approach. I was given problems to study for an upcoming exam. Are we on the correct path?
I don't see how this is going to get me the solution I was given at all????
Take sinu as t. => -dt = cosudu \[1/36 \int\limits \frac{ -dt }{ t^2 } = \frac{ 1 }{ 36t } + C = cosecu/36 + C = \frac{ \sqrt6 }{ 36\sqrt{x^2 - 6} } \]
|dw:1349017386008:dw|
LOL ITS FUNNY CUZ UR NAME IS PRECAL AND UR DOING CAL LOL
|dw:1349017310113:dw| since we let x = 6 secu secu = x/6 cosu = 6/x You can find the opposite side, hence sin u, and cscu in terms of x
gotta study all types of math. I am just more comfortable at the precal and below level
Yeaah my bad. Thats \[\sqrt x\] NOT \[\sqrt 6\]
|dw:1349017504869:dw|
Yes. And what is sin u and cscu?
|dw:1349017612360:dw|
And cscu?
|dw:1349017652671:dw|
And remember we got \[-\frac{1}{36}cscu+C\]For the integral?
|dw:1349017768354:dw|
ALTERNATE: \[x^2 - 36 = t^2 =>tdt=xdx\] \[I = \int\limits \frac{ tdt }{ t^3\sqrt{t^2 + 36}} = \int\limits \frac{ dt }{t^3 \sqrt{ 1 + \frac{ 36 }{ t^2 }} }\]
Thanks Callisto, you were correct (back to the basics)
You're welcome :)
that is in reference to drawing the triangle. Yes, I got my solution but now I have to go back and study the process
Take: \[ 1 + \frac{ 36 }{ t^2 } = u = > du = -36dt/t^3\]
You would get the same answer again after the sub. :)
@siddhantsharan I am always interested in the alternate approach, especially if it gets me to the answer quicker
thanks everyone, gotta take a break and look at this later........
Perhaps it's easier for you to read. (PS: I'm not giving direct answer!!) First Method: By Trigo Sub.\[\int \frac{dx}{(x^2 -36)^\frac{3}{2}}\]Let x=6sec u; dx = 6secu tanu du. Then, the integral becomes\[=\int \frac{6secu \ tanu\ du}{((6secu)^2 -36)^\frac{3}{2}}\]\[=\int \frac{6secu \ tanu\ du}{[36(sec^2u -1)]^\frac{3}{2}}\]\[=\int \frac{6secu \ tanu\ du}{[36(tan^2u)]^\frac{3}{2}}\]\[=\frac{1}{36}\int \frac{secu \ tanu\ du}{tan^3u}\]\[=\frac{1}{36}\int cscutanudu\]\[=-\frac{1}{36}\csc u+C\] Since x=6secu , secu = x/6 , cosu = 6/x By Pyth. Thm., opposite side is \(\sqrt{x^2-6^2}=\sqrt{x^2-36}\) So, cscu = \(\frac{x}{\sqrt{x^2-36}}\). Substituting this into the last step, you can get your answer.
Second method: by substitution, suggested by @siddhantsharan \[\int \frac{dx}{(x^2 -36)^\frac{3}{2}}\]Let x^2-36 = t^2 ; 2x dx = 2tdt => x dx = t dt => \(dx = \frac{t dt}{ x} = \frac{t dt}{ \sqrt{t^2+36}}\) PS: we already know x \(\ge\) 6, so it should be +ve sqrt The integral becomes \[\int \frac{tdt}{\sqrt{(t^2 +36)}(t^3)}\]\[=\int \frac{\frac{tdt}{t}}{\sqrt{\frac{(t^2 +36)}{t^2}}(t^3)}\] \[=\int \frac{dt}{\sqrt{1+\frac{36}{t^2}}(t^3)}\]Let u = 1+ (36/t^3) ; du = \(\frac{-72}{t^3}\)dt The integral becomes \[=-\frac{1}{72}\int \frac{du}{\sqrt{u}}\]\[=-\frac{1}{72}(2\sqrt{u}) +C\]\[=-\frac{1}{36}(\sqrt{u}) +C\] u= 1+ (36/t^2) And t^2 = x^2-36, So,\( u = 1 +\frac{36}{x^2-36} = \frac{x^2}{x^2-36}\) Sub. \( u = \frac{x^2}{x^2-36}\) into the last step, and simplify it, you should be able to get the answer. It's so amazing :D
@Callisto Clarified succinctly. Really well done :D
Thanks you are so awesome :) Both of you, wish I could give more than 1 medal

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