Integrate cos(sqrt(5t)) from x to 6

[Calculus!]

Here's the question:

integrate the function cos(sqrt(5t)) with lower bound: x and Upper bound: 6

 

[Dick]

Do a u substitution and then integrate by parts.

 

[Calculus!]

but how?

Should i integrate cos (sqrt(5t)) to 10/3 sin t ^3/2?

 

[Dick]

but how?

Should i integrate cos (sqrt(5t)) to 10/3 sin t ^3/2?

That's completely wrong. If you differentiate (10/3)*sin(t^(3/2)) you don't anything like
cos(sqrt(5t)). Try the substitution u^2=5t.

 

[Calculus!]

would du=5/2u?

because i did
u^2 = 5t
2udu = 5
du = 5/2u

 

[Dick]

What you want to find is dt in terms of du. u^2=5t -> 2u*du=5*dt. dt=(2/5)*u*du.

 

[Calculus!]

Thanks that is helpful, but i do not know how to apply it. Do you think you can walk me through the problem? I feel I have to see it before I can do it.

 

[Dick]

You are doing fine. Use the substitution to write cos(sqrt(5t))*dt completely in terms of u. It's not that hard.

 

[Calculus!]

i am just confused because the lower bound is x. I usually deal with the upper bound being x. Does this change the problem?

would i get 2/5 u cos(sqrt(5t)) integrated from 6 to x ? do i change cos to sin?

 

[Dick]

i am just confused because the lower bound is x. I usually deal with the upper bound being x. Does this change the problem?

would i get 2/5 u cos(sqrt(5t)) integrated from 6 to x ? do i change cos to sin?

Uhhhhh. Your question is making me think that you didn't post the full problem. You want the DERIVATIVE of the integral, right? Not the integral.

 

[Dick]

Here's the question:

integrate the function cos(sqrt(5t)) with lower bound: x and Upper bound: 6

If you actually want the derivative of that, pretend you know how to integrate it. So you have an F(t) such that F'(t)=cos(sqrt(5t)). By the FTOC, the integral is F(6)-F(x), right? What's the derivative of that? There is a difference between having x in the upper limit and the lower. Do you see what it is?