Question Statement
Evaluate the integral:
β«6Οβ3Οββcos(t),dt
Background and Explanation
To solve this integral, we need to recognize that the integral of cos(t) is a standard antiderivative. Specifically:
β«cos(t),dt=sin(t)+C
Thus, we will integrate cos(t) and then apply the limits of integration.
Solution
-
Recognize the Standard Integral:
The given integral is:
β«6Οβ3Οββcos(t),dt
Using the standard formula for the integral of cos(t):
β«cos(t),dt=sin(t)
-
Apply the Limits of Integration:
Now, we evaluate sin(t) at the limits 3Οβ and 6Οβ:
sin(t)β6Οβ3Οββ
This means we need to compute:
sin(3Οβ)βsin(6Οβ)
-
Compute the Sine Values:
From trigonometric values, we know:
sin(3Οβ)=23ββandsin(6Οβ)=21β
Substituting these values, we get:
23βββ21β
-
Simplify the Result:
Simplifying the expression:
23ββ1β
Therefore, the value of the integral is:
23ββ1ββ
- Integral of Cosine:
β«cos(t),dt=sin(t)
Summary of Steps
-
Recognize that β«cos(t),dt=sin(t).
-
Apply the limits of integration:
sin(3Οβ)βsin(6Οβ)
-
Use known trigonometric values:
sin(3Οβ)=23ββ,sin(6Οβ)=21β
-
Simplify the result to:
23ββ1β
-
Final answer:
23ββ1ββ