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

• Calculus!
In summary, the conversation is discussing how to integrate the function cos(sqrt(5t)) with a lower bound of x and upper bound of 6. The suggested method involves using substitution and integration by parts, but there is confusion on how to apply it. The expert clarifies the steps and points out the difference between having x in the upper limit and lower limit when finding the derivative of the integral.
Calculus!
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.

but how?

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

Calculus! said:
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.

would du=5/2u?

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?

Calculus! said:
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.

Calculus! said:
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?

## 1. What is the formula for integrating cos(sqrt(5t))?

The formula for integrating cos(sqrt(5t)) is ∫cos(sqrt(5t)) dt = (2/5)sin(sqrt(5t)) + C

## 2. How do you solve the integral of cos(sqrt(5t)) from x to 6?

To solve this integral, you first need to apply the formula for integrating cos(sqrt(5t)). Then, you can substitute the upper limit, 6, for the variable t and solve for the resulting expression. Finally, subtract the result of substituting the lower limit, x, for t from the previous result to get the final answer.

## 3. Can the integral of cos(sqrt(5t)) be solved using a different method?

Yes, there are other methods that can be used to solve this integral, such as trigonometric substitution or u-substitution. However, the formula method is often the most straightforward and efficient for this particular integral.

## 4. What happens if I substitute a negative value for t in the integral of cos(sqrt(5t))?

If you substitute a negative value for t, you will get a complex number as a result. This is because the square root of a negative number cannot be represented in real numbers. Therefore, the integral of cos(sqrt(5t)) can only be solved for non-negative values of t.

## 5. How can the integral of cos(sqrt(5t)) be applied in real-life situations?

The integral of cos(sqrt(5t)) can be used in physics and engineering to calculate the displacement or position of a body in simple harmonic motion. It can also be used in electrical engineering to analyze the behavior of circuits with oscillating currents or voltages.

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