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- anonymous

Someone please help me real quick. Im going to post the question now

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- anonymous

Someone please help me real quick. Im going to post the question now

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- anonymous

given
\[\cos(\sqrt{t})/\sqrt{t}\]
If l let u = sqt(t) then does that mean it applies to both cos funtion and root funtion, or just the inner most sqrt inside cos????????

- anonymous

both..........

- anonymous

that doesnt help me for taking the indefinite integral though

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- anonymous

because u have given new value to sqrt (t) no matter where it is

- anonymous

u might need to post your question to have opinions on that problem

- anonymous

\[\int\limits_{}^{}\cos(\sqrt{t})/\sqrt{t}\]

- anonymous

You need to consider
\[\frac{\mathbb{d}u}{\mathbb{d}x} \]

- anonymous

/dt*

- anonymous

try sqrt(t)= z

- anonymous

I was trying to get ride of the root t in the botom with \[dt = 2\sqrt{t} du\]

- anonymous

You need to substitute in u for root t there aswell..

- anonymous

after substituting z=t^1/2 u give get as below
integration sign ( cosz * 2z^2 dt )

- anonymous

then u can use integration by parts keeping z^2 as first term and cosz as second

- anonymous

let me show you what im trying

- anonymous

let u = \[\sqrt{t}\]
then \[du = 1/(2\sqrt{t})dt\]
\[dt = 2\sqrt{t}du\]
then i get
\[\int\limits_{}^{} \cos(u)/u 2\sqrt{t}du\]

- anonymous

is this wrong?

- anonymous

change t^1/2 to u because now we cant have t in new eqation, now we are dealing with u only

- anonymous

oooooooooooo

- anonymous

got it lol

- anonymous

u r on right track ..well done

- anonymous

bang !!! cheers!!! good luck with other problems too !!

- anonymous

thank you so much!

- anonymous

keep up the gud work !!!

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