Trig substitutions with integrals. Really basic question.

- anonymous

Trig substitutions with integrals. Really basic question.

- schrodinger

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- anonymous

\[\int\limits_{2\sqrt2}^{4}\frac{ 1 }{ (t^3)\sqrt(t^2-1)dt }\]

- anonymous

so, I know how to do the entire problem, but, when it comes to plugging in the cos(4) and cos(2root2) i get lost.

- anonymous

|dw:1360033290591:dw|

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## More answers

- zepdrix

You plugged everything in correctly, I think maybe you just made a tiny boo boo at the end.
See how you're squared secant that you're left with is in the `denominator`?

- anonymous

oh ya cos^2x

- zepdrix

Oh the limits? yah let's not forget that the original limits were in terms of `t`.
We can't plug those into our result.
We have 2 options.
Plug the limits into our original substitution that we made, to find new limits of integration. `Or` we can back-substitute and plug the t values in.

- anonymous

lets find t now, thats where Im lost

- anonymous

the other problem i had was real easy, root 2 and 2

- anonymous

now im not sure how to use these weird ones

- zepdrix

\[\large t=\sec \theta\]The lower limit is \(2\sqrt2\). giving us,\[\large \frac{2}{\sqrt2}=\sec \theta\]
Which is a special angle, maybe it just doesn't look like it :)\[\large \frac{2}{\sqrt2}=\frac{1}{\cos \theta} \theta \qquad \rightarrow \qquad \cos \theta = \frac{\sqrt2}{2}\]

- anonymous

|dw:1360034188203:dw|

- anonymous

ahhhh! algebra bites me in the retrice

- zepdrix

Solution looks good so far :)

- anonymous

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- zepdrix

Umm if I'm remembering my angles correctly, I think that's at \(\dfrac{\pi}{4}\).
Sine and Cosine both measure sqrt2/2 when theta is pi/4.

- zepdrix

Doesn't sound familiar? :D

- anonymous

ya

- anonymous

hmm now im lost on the 4

- zepdrix

\[\large 4=\sec \theta \qquad \rightarrow \qquad \cos \theta=1/4\]Hmmmmm..

- zepdrix

\[\large \cos \theta=\left(\frac{1}{2}\right)^2\]Hmm no I don't think that's going to get us in the right direction...

- zepdrix

\[\large \cos \theta=1/4 \qquad \rightarrow \qquad \theta=\arccos\left(\frac{1}{4}\right)\]I think we would have to leave it like this.

- anonymous

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- zepdrix

Hmm I'm not quite sure what you did there :o

- anonymous

not sure if that is legal, i multiplied both sides by sec^2/sec^2

- zepdrix

But it looks like you canceled how a `secant` and a `secant squared` somehow :o

- anonymous

ya thats dumb my bad

- anonymous

|dw:1360035089559:dw|

- zepdrix

no :c i dont think that works either.
I'm pretty sure we have to leave it as the arccosine.

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