anonymous
  • anonymous
elavualte indefinite integrals 1/x^(1/2) cos.( t^1/2 +3) dx
Calculus1
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
\[\huge \int\limits \frac{1}{x^{1/2}}\cdot \cos(t^{1/2}+3) \; dx\]Like this? Is that t suppose to be an x by chance?
anonymous
  • anonymous
yes, sorry. that is one t^1/2 not x^1/2
anonymous
  • anonymous
*that one is

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

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