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Mathematics 19 Online
OpenStudy (anonymous):

For the function f(t) = (cos√t )/(√t) , find the antiderivative F of f that satisfies the given condition F(pi^2 / 4) = 5.

OpenStudy (freckles):

You could perform a substitution

OpenStudy (freckles):

I'm confused...Why did you post a link to this page?

OpenStudy (freckles):

\[\int\limits_{}^{}\frac{\cos(\sqrt{t})}{\sqrt{t}} dt \] do u=sqrt(t)

OpenStudy (anonymous):

never mind, can u just help me?

OpenStudy (anonymous):

yes its cosine of sqrt of t

OpenStudy (freckles):

let me know if you need more help after trying that sub

OpenStudy (anonymous):

can you pls help I have no clue how to do this. please?

OpenStudy (anonymous):

@freckles

OpenStudy (freckles):

\[u=\sqrt{t} \\ \frac{du}{dt}=?\]

OpenStudy (freckles):

do you know how to find the derivative of sqrt(t)?

OpenStudy (freckles):

and remember that is just t^(1/2)

OpenStudy (freckles):

use power rule

OpenStudy (anonymous):

derivative would be 1/(2√t )right?

OpenStudy (freckles):

yes

OpenStudy (freckles):

\[u=\sqrt{t} \\ 2 du =\frac{1}{\sqrt{t}} dt \] tried to use this substitution in your integral

OpenStudy (anonymous):

okay! I get that part now that you showed, but then what?

OpenStudy (anonymous):

i know you have to take the integral but idk integral of what

OpenStudy (freckles):

\[F(t)=\int\limits \frac{\cos(\sqrt{t})}{\sqrt{t}} dt \] replace the sqrt(t) inside the cosine with u replace the 1/sqrt(t) dt with 2 du I'm asking you to use the sub

OpenStudy (freckles):

\[F(t) =\int\limits \cos(\sqrt{t}) \frac{1}{\sqrt{t}} dt \] \[F(t)=\int\limits \cos(u) 2 du =2 \int\limits \cos(u) du\] Can you integrate cos(u) w.r.t. u ?

OpenStudy (anonymous):

no.. can u just help me all the way through this problm? i have a hw page full of these type of questions I promise I'll do them all!

OpenStudy (freckles):

so you don't know what function you can take derivative of that will give you cos?

OpenStudy (freckles):

\[( ? )'=\cos(u)\]

OpenStudy (anonymous):

sine

OpenStudy (freckles):

yes

OpenStudy (anonymous):

sorry i didnt know what u meant

OpenStudy (freckles):

so F(t)=2sin(u)+C where u=sqrt(t)

OpenStudy (anonymous):

ok and then?

OpenStudy (anonymous):

so thats the antiderivative right?

OpenStudy (anonymous):

but idk how to satisfy the other equation!

OpenStudy (freckles):

Well you have an unknown constant C there

OpenStudy (freckles):

use the condition to find it

OpenStudy (freckles):

F(a)=b means you will replace t with a and F(a) with b

OpenStudy (anonymous):

can you show me?

OpenStudy (freckles):

let me see you try to plug in the values

OpenStudy (freckles):

replace t with pi^2/4 and F(pi^2/4) with 5 and solve for C

OpenStudy (anonymous):

can you show m actually like you did above w/ pictures? not just words becuz its hard for me to uderstand

OpenStudy (freckles):

You aren't willing to try to apply what I said?

OpenStudy (anonymous):

im sorry i just... its not that im not willing i have to see it to understand it.

OpenStudy (freckles):

Do you know what we found for F(t)?

OpenStudy (anonymous):

so 2sin√(pi^2/4) +c = what?

OpenStudy (freckles):

can you tell me the function so far?

OpenStudy (anonymous):

|dw:1424455883300:dw|

OpenStudy (freckles):

well don't forget what that expression equals \[F(t)= 2 \sin(\sqrt{t})+C\]

OpenStudy (anonymous):

so then?

OpenStudy (freckles):

you are going to replace all the t's with pi^2/4\[F(\frac{\pi^2}{4})=2 \sin(\sqrt{\frac{\pi^2}{4}})+C\]

OpenStudy (freckles):

And you are given what F(pi^2/4) equals

OpenStudy (freckles):

so replace F(pi^2/4) with 5 since F(pi^2/4)=5

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