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anonymous
 4 years ago
elavualte indefinite integrals
1/x^(1/2) cos.( t^1/2 +3) dx
anonymous
 4 years ago
elavualte indefinite integrals 1/x^(1/2) cos.( t^1/2 +3) dx

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zepdrix
 4 years ago
Best 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?

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

zepdrix
 4 years ago
Best 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.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1So 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.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Were you able to get a dU? Or confused? :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but trying to figure it out

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1dw: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.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for du, we just eed to take the derivative of 1/t^1/2?

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

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1From 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. \[\large \frac{du}{dt}=\frac{1}{2t^{1/2}} \qquad \rightarrow \qquad \large du=\frac{1}{2t^{1/2}}dt\]

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1We'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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can write it as square root of t?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1oh 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do I do with it?

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

zepdrix
 4 years ago
Best 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,\[\large \int\limits \cos(t^{1/2}+3)\left(\frac{1}{t^{1/2}}dt\right) \qquad \rightarrow \qquad \int\limits \cos(u)(2du)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355688113175:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the second step , do I have to write the du?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1If 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1sure 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure, i need to have lunch too. ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355688628735:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i was trying to simplify it. is it correct?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1yah looks good so far c:

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355689271796:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i always get confused when calculating the du

zepdrix
 4 years ago
Best 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

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

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1No, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u= x^ 2 1 , then du = 2x

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1yah 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..?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can i use the wolframalpha and see the answer ? :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i know i will get the answer there, but I need to understand how to do it.

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1http://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 :\

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1"stepbystep solution" button *

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the solution is right in front of me step by step .hmm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i will go by that rule only,
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