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dfresenius
Trig substitutions with integrals. Really basic question.
\[\int\limits_{2\sqrt2}^{4}\frac{ 1 }{ (t^3)\sqrt(t^2-1)dt }\]
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.
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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`?
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.
lets find t now, thats where Im lost
the other problem i had was real easy, root 2 and 2
now im not sure how to use these weird ones
\[\large t=\sec \theta\]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 :)\[\large \frac{2}{\sqrt2}=\frac{1}{\cos \theta} \theta \qquad \rightarrow \qquad \cos \theta = \frac{\sqrt2}{2}\]
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ahhhh! algebra bites me in the retrice
Solution looks good so far :)
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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.
Doesn't sound familiar? :D
hmm now im lost on the 4
\[\large 4=\sec \theta \qquad \rightarrow \qquad \cos \theta=1/4\]Hmmmmm..
\[\large \cos \theta=\left(\frac{1}{2}\right)^2\]Hmm no I don't think that's going to get us in the right direction...
\[\large \cos \theta=1/4 \qquad \rightarrow \qquad \theta=\arccos\left(\frac{1}{4}\right)\]I think we would have to leave it like this.
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Hmm I'm not quite sure what you did there :o
not sure if that is legal, i multiplied both sides by sec^2/sec^2
But it looks like you canceled how a `secant` and a `secant squared` somehow :o
ya thats dumb my bad
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no :c i dont think that works either. I'm pretty sure we have to leave it as the arccosine.