The given integral is:

$I = \int_0^{\frac{\pi}{2}} \sin^6{x} \cos^8{x} \, dx$

We will solve this step by step.

Step 1: Beta Function Representation

The Beta function $B(m, n)$ is defined as:

$B(m, n) = 2 \int_0^{\frac{\pi}{2}} \sin^{2m-1}(x) \cos^{2n-1}(x) \, dx$

For $\sin^a{x} \cos^b{x}$, this formula applies when powers are rewritten as $2m-1$ and $2n-1$. Let's match this integral:

• $\sin^6{x} \cos^8{x}$
  • Rewrite as $\sin^{2 \cdot 3}(x) \cos^{2 \cdot 4}(x)$
  • Thus, $2m-1 = 7$ (so $m = 4$) and $2n-1 = 9$ (so $n = 5$).

Step 2: Using Beta Function

The integral can be rewritten as:

$I = \frac{1}{2} B\left( \frac{7+1}{2}, \frac{9+1}{2} \right)$

Simplify the arguments of $B$:

• $\frac{7+1}{2} = 4$
• $\frac{9+1}{2} = 5$

Therefore:

$I = \frac{1}{2} B(4, 5)$

Step 3: Beta Function and Gamma Function

The Beta function $B(m, n)$ is related to the Gamma function $\Gamma(x)$ as:

$B(m, n) = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$

For $B(4, 5)$:

• $\Gamma(4) = (4-1)! = 3! = 6$
• $\Gamma(5) = (5-1)! = 4! = 24$
• $\Gamma(4+5) = \Gamma(9) = (9-1)! = 8! = 40320$

Now compute $B(4, 5)$:

$B(4, 5) = \frac{\Gamma(4) \Gamma(5)}{\Gamma(9)} = \frac{6 \cdot 24}{40320}$

Simplify:

$B(4, 5) = \frac{144}{40320} = \frac{1}{280}$

Step 4: Final Result

Substitute $B(4, 5) = \frac{1}{280}$ into $I$:

$I = \frac{1}{2} B(4, 5) = \frac{1}{2} \cdot \frac{1}{280} = \frac{1}{560}$

Final Answer:

$\int_0^{\frac{\pi}{2}} \sin^6{x} \cos^8{x} \, dx = \frac{1}{560}$

Do you want a detailed explanation for any step? Let me know!

Here are 5 related questions to deepen your understanding:

1. How do you derive the Beta function formula from the integral of $\sin$ and $\cos$ powers?
2. What is the relationship between the Gamma and factorial functions?
3. How can the Beta function be extended to other integrals?
4. Solve $\int_0^{\frac{\pi}{2}} \sin^4{x} \cos^6{x} \, dx$.
5. What are the properties of the Beta function and its symmetry?

Tip: The Beta function is highly useful in solving integrals of powers of trigonometric functions over $[0, \frac{\pi}{2}]$. It connects directly to the Gamma function for factorial simplifications.