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|dw:1349015981615:dw|

This looks like one of the trig substitutions.

or maybe integration by parts would work.

|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 :)

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"

cos u/ sin^2 u = cscu cotu
\[\int \csc u \cot udu = -cscu +C\]
Time to draw a triangle...

I don't see how this is going to get me the solution I was given at all????

|dw:1349017386008:dw|

LOL ITS FUNNY CUZ UR NAME IS PRECAL AND UR DOING CAL LOL

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|

Thanks Callisto, you were correct (back to the basics)

You're welcome :)

Take:
\[ 1 + \frac{ 36 }{ t^2 } = u = > du = -36dt/t^3\]

You would get the same answer again after the sub. :)

thanks everyone, gotta take a break and look at this later........

Thanks you are so awesome :) Both of you, wish I could give more than 1 medal