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

Calculus1
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\[\huge \int\limits \frac{1}{x^{1/2}}\cdot \cos(t^{1/2}+3) \; dx\]Like this? Is that t suppose to be an x by chance?
yes, sorry. that is one t^1/2 not x^1/2
*that one is

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\[\large \int\limits\limits \frac{1}{t^{1/2}}\cdot \cos(t^{1/2}+3) \; dt\] Oh ok :) Hmm looks like you can do a nice U-substitution for this one.
okay,
So with a U-sub, 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.
Were you able to get a dU? Or confused? :)
kind of confused
but trying to figure it out
|dw:1355686904344:dw|We'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.
If we let,\[\large u=t^{1/2}+3\]Taking the derivative with respect to t gives us,\[\large \frac{du}{dt}=\frac{1}{2t^{1/2}}\]
for du, we just eed to take the derivative of 1/t^1/2?
No, we want to take the derivative of the other part :O See what I did with u?
yes
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. \[\large \frac{du}{dt}=\frac{1}{2t^{1/2}} \qquad \rightarrow \qquad \large du=\frac{1}{2t^{1/2}}dt\]
okay,
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?
can write it as square root of t?
not sure
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?
yeah
what do I do with it?
|dw:1355687571473:dw|
We'll multiply both sides by 2, to get rid of the 1/2.\[\large 2du=\frac{1}{t^{1/2}}dt\]
okay,
\[\large u=t^{1/2}+3 \qquad \qquad 2du=\frac{1}{t^{1/2}}dt\]Plugging these into our integral gives us,\[\large \int\limits \cos(t^{1/2}+3)\left(\frac{1}{t^{1/2}}dt\right) \qquad \rightarrow \qquad \int\limits \cos(u)(2du)\]
okay. i see
2 sin(t^1/2 + 3)?
Yesss very good.
|dw:1355688113175:dw|
in the second step , do I have to write the du?
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.
oh okay
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.
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
sure, i need to have lunch too. ...
hah XD
|dw:1355688628735:dw|
i was trying to simplify it. is it correct?
yah looks good so far c:
Hmm I don't think this one will work out nicely with a U-sub. Have you learned about Trig-sbustitutions yet?
|dw:1355689271796:dw|
no, not yet
Hmm
i always get confused when calculating the du
\[\large u=x^2-1 \qquad \rightarrow \qquad du=2x \; dx\]It's just the derivative! :) Hmm I don't think this is going to help us though :( Hmmmm Thinking :3
so the one du that i have calculated is correct for the above problem?
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.
ohh...
u= x^ 2 -1 , then du = 2x
sorry
2xdx
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..?
yeah
can i use the wolframalpha and see the answer ? :)
i know i will get the answer there, but I need to understand how to do it.
http://www.wolframalpha.com/input/?i=integrate+%28x%5E2-1%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 :\
"step-by-step solution" button *
okay,
we did not learn the trig substitution yet and if I do that problem using the trig substitution I will get a 0 :)
the solution is right in front of me step by step .hmm
i will go by that rule only,

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