Math Problem Statement
. Evaluate the following definite integrals using beta-gamma functions. i. ∫ sin6 𝜃 𝑑𝜃
Solution
To evaluate the integral , we will use a connection between this type of integral and the Beta function. The Beta function is related to integrals of the form , which is the general case for powers of sine.
Step-by-Step Solution
The definite integral we need to evaluate is:
We can express this using the Beta function by recalling the following relation:
This relation holds for even powers of , where in our case (since ).
Thus, we can apply the formula as follows:
Calculating the Gamma functions:
- (this is a well-known value).
- , which simplifies to .
Using the property of the Gamma function, , we can compute as follows:
Now we substitute these values:
Thus, the integral becomes:
Final Answer:
Would you like to see more details on Beta-Gamma function properties or have any questions?
Here are some related questions for further practice:
- Evaluate .
- Evaluate .
- Use the Beta function to solve .
- How is the Beta function related to the Gamma function?
- What are the properties of the Gamma function for half-integer values?
Tip: Knowing the recurrence relation for Gamma functions can simplify complicated calculations, especially for half-integer values.
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Math Problem Analysis
Mathematical Concepts
Calculus
Definite Integrals
Beta Function
Gamma Function
Formulas
∫_0^π sin^(2a) θ dθ = (Γ(a + 1/2) Γ(1/2)) / Γ(a + 1)
Γ(x + 1) = xΓ(x)
Theorems
Properties of the Beta Function
Gamma Function Recurrence Relation
Suitable Grade Level
Undergraduate
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