Here's the question:
integrate the function cos(sqrt(5t)) with lower bound: x and Upper bound: 6
Do a u substitution and then integrate by parts.
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.
because i did
u^2 = 5t
2udu = 5
du = 5/2u
What you want to find is dt in terms of du. u^2=5t -> 2u*du=5*dt. dt=(2/5)*u*du.
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.
You are doing fine. Use the substitution to write cos(sqrt(5t))*dt completely in terms of u. It's not that hard.
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.
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?
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