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dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{2\sqrt2}^{4}\frac{ 1 }{ (t^3)\sqrt(t^21)dt }\]

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0so, I know how to do the entire problem, but, when it comes to plugging in the cos(4) and cos(2root2) i get lost.

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360033290591:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0You 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`?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Oh 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 backsubstitute and plug the t values in.

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0lets find t now, thats where Im lost

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0the other problem i had was real easy, root 2 and 2

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0now im not sure how to use these weird ones

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large t=\sec \theta\]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 :)\[\large \frac{2}{\sqrt2}=\frac{1}{\cos \theta} \theta \qquad \rightarrow \qquad \cos \theta = \frac{\sqrt2}{2}\]

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360034188203:dw

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0ahhhh! algebra bites me in the retrice

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Solution looks good so far :)

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360034363382:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Umm 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
 2 years ago
Best ResponseYou've already chosen the best response.0Doesn't sound familiar? :D

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0hmm now im lost on the 4

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large 4=\sec \theta \qquad \rightarrow \qquad \cos \theta=1/4\]Hmmmmm..

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large \cos \theta=\left(\frac{1}{2}\right)^2\]Hmm no I don't think that's going to get us in the right direction...

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large \cos \theta=1/4 \qquad \rightarrow \qquad \theta=\arccos\left(\frac{1}{4}\right)\]I think we would have to leave it like this.

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360034799626:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Hmm I'm not quite sure what you did there :o

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0not sure if that is legal, i multiplied both sides by sec^2/sec^2

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0But it looks like you canceled how a `secant` and a `secant squared` somehow :o

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0ya thats dumb my bad

dfresenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360035089559:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0no :c i dont think that works either. I'm pretty sure we have to leave it as the arccosine.
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