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zepdrixBest ResponseYou've already chosen the best response.1
\[\huge \int\limits \frac{1}{x^{1/2}}\cdot \cos(t^{1/2}+3) \; dx\]Like this? Is that t suppose to be an x by chance?
 one year ago

roselinBest ResponseYou've already chosen the best response.0
yes, sorry. that is one t^1/2 not x^1/2
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large \int\limits\limits \frac{1}{t^{1/2}}\cdot \cos(t^{1/2}+3) \; dt\] Oh ok :) Hmm looks like you can do a nice Usubstitution for this one.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
So with a Usub, we're trying to find a U and U' somewhere in the problem. If you look at the inside of the cosine, taking the derivative of that should give you something similar to the other term. So try letting,\[\large u=t^{1/2}+3\]And see if you can match up your dU from there.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Were you able to get a dU? Or confused? :)
 one year ago

roselinBest ResponseYou've already chosen the best response.0
but trying to figure it out
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
dw:1355686904344:dwWe're trying to do something like this, Simplify the integral so it's very easy to integrate. We just have to figure out what DU is so we can make sure we didn't miss anything.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
If we let,\[\large u=t^{1/2}+3\]Taking the derivative with respect to t gives us,\[\large \frac{du}{dt}=\frac{1}{2t^{1/2}}\]
 one year ago

roselinBest ResponseYou've already chosen the best response.0
for du, we just eed to take the derivative of 1/t^1/2?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
No, we want to take the derivative of the other part :O See what I did with u?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
From there, there is a process that allows us to write the dt on the other side. You can simply think of it as multiplication for your purposes. \[\large \frac{du}{dt}=\frac{1}{2t^{1/2}} \qquad \rightarrow \qquad \large du=\frac{1}{2t^{1/2}}dt\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
We're trying to replace this big chunk of stuff, Think of THIS as your variable that you're trying to solve for,\[\large \frac{1}{t^{1/2}}dt\] But we have a 1/2 in attached to it, how do we fix that?
 one year ago

roselinBest ResponseYou've already chosen the best response.0
can write it as square root of t?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
oh i don't mean that 1/2. I mean the 2 in front of the t,\[\large \frac{1}{2t^{1/2}}\]See how this doesn't quite match what I posted before?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
We'll multiply both sides by 2, to get rid of the 1/2.\[\large 2du=\frac{1}{t^{1/2}}dt\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large u=t^{1/2}+3 \qquad \qquad 2du=\frac{1}{t^{1/2}}dt\]Plugging these into our integral gives us,\[\large \int\limits \cos(t^{1/2}+3)\left(\frac{1}{t^{1/2}}dt\right) \qquad \rightarrow \qquad \int\limits \cos(u)(2du)\]
 one year ago

roselinBest ResponseYou've already chosen the best response.0
in the second step , do I have to write the du?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
If you still have the Integral bar, then yes. When the integral bar disappears, the du disappears also. It's a part of the integration process.
 one year ago

roselinBest ResponseYou've already chosen the best response.0
i have another problem, which I am trying to solve, i will do my best to try to solve it, will you guide me through that, i will only ask you to check whether it is correct or not.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
sure c: Imma go make some lunch real quick. So i brb. You can post some stuff though, and I'll take a look in a few minutes when i get back :D
 one year ago

roselinBest ResponseYou've already chosen the best response.0
sure, i need to have lunch too. ...
 one year ago

roselinBest ResponseYou've already chosen the best response.0
i was trying to simplify it. is it correct?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
yah looks good so far c:
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hmm I don't think this one will work out nicely with a Usub. Have you learned about Trigsbustitutions yet?
 one year ago

roselinBest ResponseYou've already chosen the best response.0
i always get confused when calculating the du
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large u=x^21 \qquad \rightarrow \qquad du=2x \; dx\]It's just the derivative! :) Hmm I don't think this is going to help us though :( Hmmmm Thinking :3
 one year ago

roselinBest ResponseYou've already chosen the best response.0
so the one du that i have calculated is correct for the above problem?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
No, I'm not really sure what you did there. You picked something for your U. Then instead of taking the derivative of U, you took the derivative of x^4.
 one year ago

roselinBest ResponseYou've already chosen the best response.0
u= x^ 2 1 , then du = 2x
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
yah good times :) Unfortunately, I don't think this is going to help us :( We don't have any xdx in our problem, we have 1/x^4 dx but that's not the same thing. I think you NEED a trig sub for this one.. I could be wrong though D: This is a homework problem..?
 one year ago

roselinBest ResponseYou've already chosen the best response.0
can i use the wolframalpha and see the answer ? :)
 one year ago

roselinBest ResponseYou've already chosen the best response.0
i know i will get the answer there, but I need to understand how to do it.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=integrate+%28x%5E21%29%5E%281%2F2%29%2Fx%5E4+dx Yah if you click the "show steps" button, you'll see that wolfram uses a trig sub to solve it. There might be another way, like by parts or something.. But I dunno :\
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
"stepbystep solution" button *
 one year ago

roselinBest ResponseYou've already chosen the best response.0
we did not learn the trig substitution yet and if I do that problem using the trig substitution I will get a 0 :)
 one year ago

roselinBest ResponseYou've already chosen the best response.0
the solution is right in front of me step by step .hmm
 one year ago

roselinBest ResponseYou've already chosen the best response.0
i will go by that rule only,
 one year ago
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