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precal
 4 years ago
Integration problem
precal
 4 years ago
Integration problem

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bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0This looks like one of the trig substitutions.

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0or maybe integration by parts would work.

precal
 4 years ago
Best ResponseYou've already chosen the best response.0so I will rewrite it asdw:1349016056313:dw

precal
 4 years ago
Best ResponseYou've already chosen the best response.0Am I doing it correct so far?

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0I think it's best if we leave it inside. Something tells me that will have to stay as u^(3/2)

precal
 4 years ago
Best ResponseYou've already chosen the best response.0then I don't see with trig sub I am suppose to use

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nope . Use Inverse sec :) or Sine? I forgot what to use but simple as that :D

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0It is trig, usub won't work. Yeah keep it out, it's inverse sin of 6x i believe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its nerp it is sin inverse x/6 +C :)

precal
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349016274224:dw is what my solution is suppose to be. Of course I am more interested in how to get to the solution

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.5I think we should use sec instead of sin?! Let x = 6sec (theta)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.036Sec^2(theta)  36 = 36 tan^2(theta)

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0so the bottom is: sqrt (36 tan^2(theta)) = 6tan^2(theta), yea callisto's right

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0and top is: dx = 6Sec(theta)Tan(theta)dtheta

precal
 4 years ago
Best ResponseYou've already chosen the best response.0so is my given solution incorrect?

precal
 4 years ago
Best ResponseYou've already chosen the best response.0I have some integrals at the backof my book "Integral involving x^2a^2; a>0"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \frac{ 6secutanudu }{ (36\sec^2u  36)^{3/2} } = \int\limits \frac{ 6secu tanu du }{ 6^3 \tan^3u }\] \[1/36 \int\limits secdu/\tan^2u \] \[1/36 \int\limits \frac{ cosudu }{ \sin^2u? }\]

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.5cos u/ sin^2 u = cscu cotu \[\int \csc u \cot udu = cscu +C\] Time to draw a triangle...

precal
 4 years ago
Best ResponseYou've already chosen the best response.0time 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?

precal
 4 years ago
Best ResponseYou've already chosen the best response.0I don't see how this is going to get me the solution I was given at all????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take sinu as t. => dt = cosudu \[1/36 \int\limits \frac{ dt }{ t^2 } = \frac{ 1 }{ 36t } + C = cosecu/36 + C = \frac{ \sqrt6 }{ 36\sqrt{x^2  6} } \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL ITS FUNNY CUZ UR NAME IS PRECAL AND UR DOING CAL LOL

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.5dw: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

precal
 4 years ago
Best ResponseYou've already chosen the best response.0gotta study all types of math. I am just more comfortable at the precal and below level

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeaah my bad. Thats \[\sqrt x\] NOT \[\sqrt 6\]

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.5Yes. And what is sin u and cscu?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.5And remember we got \[\frac{1}{36}cscu+C\]For the integral?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ALTERNATE: \[x^2  36 = t^2 =>tdt=xdx\] \[I = \int\limits \frac{ tdt }{ t^3\sqrt{t^2 + 36}} = \int\limits \frac{ dt }{t^3 \sqrt{ 1 + \frac{ 36 }{ t^2 }} }\]

precal
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks Callisto, you were correct (back to the basics)

precal
 4 years ago
Best ResponseYou've already chosen the best response.0that is in reference to drawing the triangle. Yes, I got my solution but now I have to go back and study the process

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take: \[ 1 + \frac{ 36 }{ t^2 } = u = > du = 36dt/t^3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You would get the same answer again after the sub. :)

precal
 4 years ago
Best ResponseYou've already chosen the best response.0@siddhantsharan I am always interested in the alternate approach, especially if it gets me to the answer quicker

precal
 4 years ago
Best ResponseYou've already chosen the best response.0thanks everyone, gotta take a break and look at this later........

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.5Perhaps 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 \(\sqrt{x^26^2}=\sqrt{x^236}\) So, cscu = \(\frac{x}{\sqrt{x^236}}\). Substituting this into the last step, you can get your answer.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.5Second method: by substitution, suggested by @siddhantsharan \[\int \frac{dx}{(x^2 36)^\frac{3}{2}}\]Let x^236 = 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^236, So,\( u = 1 +\frac{36}{x^236} = \frac{x^2}{x^236}\) Sub. \( u = \frac{x^2}{x^236}\) into the last step, and simplify it, you should be able to get the answer. It's so amazing :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Callisto Clarified succinctly. Really well done :D

precal
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks you are so awesome :) Both of you, wish I could give more than 1 medal
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