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

- anonymous

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

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- schrodinger

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- 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

yes, sorry. that is one t^1/2 not x^1/2

- anonymous

*that one is

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## More answers

- 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

okay,

- 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

Were you able to get a dU? Or confused? :)

- anonymous

kind of confused

- anonymous

but trying to figure it out

- 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

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

for du, we just eed to take the derivative of 1/t^1/2?

- zepdrix

No, we want to take the derivative of the other part :O See what I did with u?

- anonymous

yes

- 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

okay,

- 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

can write it as square root of t?

- anonymous

not sure

- 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

yeah

- anonymous

what do I do with it?

- zepdrix

|dw:1355687571473:dw|

- zepdrix

We'll multiply both sides by 2, to get rid of the 1/2.\[\large 2du=\frac{1}{t^{1/2}}dt\]

- anonymous

okay,

- 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

okay. i see

- anonymous

2 sin(t^1/2 + 3)?

- zepdrix

Yesss very good.

- anonymous

|dw:1355688113175:dw|

- anonymous

in the second step , do I have to write the du?

- 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

oh okay

- 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

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

sure, i need to have lunch too. ...

- zepdrix

hah XD

- anonymous

|dw:1355688628735:dw|

- anonymous

i was trying to simplify it. is it correct?

- zepdrix

yah looks good so far c:

- zepdrix

Hmm I don't think this one will work out nicely with a U-sub. Have you learned about Trig-sbustitutions yet?

- anonymous

|dw:1355689271796:dw|

- anonymous

no, not yet

- zepdrix

Hmm

- anonymous

i always get confused when calculating the du

- 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

so the one du that i have calculated is correct for the above problem?

- 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

ohh...

- anonymous

u= x^ 2 -1 , then du = 2x

- anonymous

sorry

- anonymous

2xdx

- 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

yeah

- anonymous

can i use the wolframalpha and see the answer ? :)

- anonymous

i know i will get the answer there, but I need to understand how to do it.

- 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

"step-by-step solution" button *

- anonymous

okay,

- 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

the solution is right in front of me step by step .hmm

- anonymous

i will go by that rule only,

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