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

dfreseniusBest ResponseYou've already chosen the best response.0
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.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1360033290591:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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`?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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 backsubstitute and plug the t values in.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
lets find t now, thats where Im lost
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
the other problem i had was real easy, root 2 and 2
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
now im not sure how to use these weird ones
 one year ago

zepdrixBest 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 :)\[\large \frac{2}{\sqrt2}=\frac{1}{\cos \theta} \theta \qquad \rightarrow \qquad \cos \theta = \frac{\sqrt2}{2}\]
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1360034188203:dw
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ahhhh! algebra bites me in the retrice
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Solution looks good so far :)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1360034363382:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Doesn't sound familiar? :D
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
hmm now im lost on the 4
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large 4=\sec \theta \qquad \rightarrow \qquad \cos \theta=1/4\]Hmmmmm..
 one year ago

zepdrixBest 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...
 one year ago

zepdrixBest 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.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1360034799626:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Hmm I'm not quite sure what you did there :o
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.0
But it looks like you canceled how a `secant` and a `secant squared` somehow :o
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ya thats dumb my bad
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1360035089559:dw
 one year ago

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