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precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349015981615:dw
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
This looks like one of the trig substitutions.
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
or maybe integration by parts would work.
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349016003653:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
so I will rewrite it asdw:1349016056313:dw
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
Am I doing it correct so far?
 one year ago

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

precal Group TitleBest ResponseYou've already chosen the best response.0
then I don't see with trig sub I am suppose to use
 one year ago

justine027 Group TitleBest ResponseYou've already chosen the best response.0
Nope . Use Inverse sec :) or Sine? I forgot what to use but simple as that :D
 one year ago

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

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349016200316:dw
 one year ago

justine027 Group TitleBest ResponseYou've already chosen the best response.0
Its nerp it is sin inverse x/6 +C :)
 one year ago

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

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

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
36Sec^2(theta)  36 = 36 tan^2(theta)
 one year ago

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

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
and top is: dx = 6Sec(theta)Tan(theta)dtheta
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
so is my given solution incorrect?
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
yea precal
 one year ago

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

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits \frac{ 6secutanudu }{ (36\sec^2u  36)^{3/2} } = \int\limits \frac{ 6secu tanu du }{ 6^3 \tan^3u }\] \[1/36 \int\limits secdu/\tan^2u \] \[1/36 \int\limits \frac{ cosudu }{ \sin^2u? }\]
 one year ago

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

precal Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

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

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.1
Take sinu as t. => dt = cosudu \[1/36 \int\limits \frac{ dt }{ t^2 } = \frac{ 1 }{ 36t } + C = cosecu/36 + C = \frac{ \sqrt6 }{ 36\sqrt{x^2  6} } \]
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349017386008:dw
 one year ago

alanli123 Group TitleBest ResponseYou've already chosen the best response.0
LOL ITS FUNNY CUZ UR NAME IS PRECAL AND UR DOING CAL LOL
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
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
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
gotta study all types of math. I am just more comfortable at the precal and below level
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.1
Yeaah my bad. Thats \[\sqrt x\] NOT \[\sqrt 6\]
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349017504869:dw
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Yes. And what is sin u and cscu?
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349017612360:dw
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
And cscu?
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349017652671:dw
 one year ago

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

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1349017768354:dw
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.1
ALTERNATE: \[x^2  36 = t^2 =>tdt=xdx\] \[I = \int\limits \frac{ tdt }{ t^3\sqrt{t^2 + 36}} = \int\limits \frac{ dt }{t^3 \sqrt{ 1 + \frac{ 36 }{ t^2 }} }\]
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
Thanks Callisto, you were correct (back to the basics)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
You're welcome :)
 one year ago

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

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.1
Take: \[ 1 + \frac{ 36 }{ t^2 } = u = > du = 36dt/t^3\]
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.1
You would get the same answer again after the sub. :)
 one year ago

precal Group TitleBest 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
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
thanks everyone, gotta take a break and look at this later........
 one year ago

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

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Second 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
 one year ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.1
@Callisto Clarified succinctly. Really well done :D
 one year ago

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