Math Problem Statement

. Evaluate the following definite integrals using beta-gamma functions. i. ∫ sin6 𝜃 𝑑𝜃

Solution

To evaluate the integral sin6θdθ\int \sin^6 \theta \, d\theta, we will use a connection between this type of integral and the Beta function. The Beta function B(x,y)B(x, y) is related to integrals of the form 0πsin2x1θdθ\int_0^\pi \sin^{2x-1} \theta \, d\theta, which is the general case for powers of sine.

Step-by-Step Solution

The definite integral we need to evaluate is:

I=0πsin6θdθI = \int_0^\pi \sin^6 \theta \, d\theta

We can express this using the Beta function by recalling the following relation:

0πsin2aθdθ=Γ(a+12)Γ(12)Γ(a+1)\int_0^\pi \sin^{2a} \theta \, d\theta = \frac{\Gamma(a + \frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(a + 1)}

This relation holds for even powers of sinθ\sin \theta, where a=3a = 3 in our case (since sin6θ=sin2(3)θ\sin^6 \theta = \sin^{2(3)} \theta).

Thus, we can apply the formula as follows:

I=Γ(3+12)Γ(12)Γ(4)I = \frac{\Gamma(3 + \frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(4)}

Calculating the Gamma functions:

  • Γ(3+12)=Γ(72)\Gamma(3 + \frac{1}{2}) = \Gamma\left(\frac{7}{2}\right)
  • Γ(12)=π\Gamma(\frac{1}{2}) = \sqrt{\pi} (this is a well-known value).
  • Γ(4)=3!\Gamma(4) = 3!, which simplifies to 66.

Using the property of the Gamma function, Γ(x+1)=xΓ(x)\Gamma(x + 1) = x\Gamma(x), we can compute Γ(72)\Gamma\left(\frac{7}{2}\right) as follows:

Γ(72)=52Γ(52)\Gamma\left(\frac{7}{2}\right) = \frac{5}{2} \Gamma\left(\frac{5}{2}\right) Γ(52)=32Γ(32)\Gamma\left(\frac{5}{2}\right) = \frac{3}{2} \Gamma\left(\frac{3}{2}\right) Γ(32)=12Γ(12)=12π\Gamma\left(\frac{3}{2}\right) = \frac{1}{2} \Gamma\left(\frac{1}{2}\right) = \frac{1}{2} \sqrt{\pi}

Now we substitute these values:

Γ(72)=523212π=158π\Gamma\left(\frac{7}{2}\right) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}

Thus, the integral becomes:

I=158ππ6=15π48=5π16I = \frac{\frac{15}{8} \sqrt{\pi} \cdot \sqrt{\pi}}{6} = \frac{15 \pi}{48} = \frac{5 \pi}{16}

Final Answer:

0πsin6θdθ=5π16\int_0^\pi \sin^6 \theta \, d\theta = \frac{5 \pi}{16}

Would you like to see more details on Beta-Gamma function properties or have any questions?

Here are some related questions for further practice:

  1. Evaluate 0πsin4θdθ\int_0^\pi \sin^4 \theta \, d\theta.
  2. Evaluate 0πcos6θdθ\int_0^\pi \cos^6 \theta \, d\theta.
  3. Use the Beta function to solve 01(1x2)2dx\int_0^1 (1 - x^2)^2 dx.
  4. How is the Beta function related to the Gamma function?
  5. 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