1. **Problem:** Evaluate the integral $$\int_0^{\frac{\pi}{2}} \cos^2 x \sin^6 x \, dx$$. 2. **Formula and rules:** We use the Beta function relation for integrals of the form $$\int_0^{\frac{\pi}{2}} \sin^{m} x \cos^{n} x \, dx = \frac{1}{2} B\left(\frac{m+1}{2}, \frac{n+1}{2}\right)$$ where $$B(p,q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$ and $$\Gamma$$ is the Gamma function. 3. **Apply to our integral:** Here, $$m=6$$ and $$n=2$$, so $$\int_0^{\frac{\pi}{2}} \sin^6 x \cos^2 x \, dx = \frac{1}{2} B\left(\frac{6+1}{2}, \frac{2+1}{2}\right) = \frac{1}{2} B\left(\frac{7}{2}, \frac{3}{2}\right).$$ 4. **Evaluate Beta function:** Using the Gamma function properties, $$B\left(\frac{7}{2}, \frac{3}{2}\right) = \frac{\Gamma\left(\frac{7}{2}\right) \Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{7}{2} + \frac{3}{2}\right)} = \frac{\Gamma\left(\frac{7}{2}\right) \Gamma\left(\frac{3}{2}\right)}{\Gamma(5)}.$$ Recall: - $$\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$ - $$\Gamma\left(n + \frac{1}{2}\right) = \frac{(2n)!}{4^n n!} \sqrt{\pi}$$ for integer $$n$$. Calculate each: - $$\Gamma\left(\frac{3}{2}\right) = \frac{1}{2} \sqrt{\pi}$$ - $$\Gamma\left(\frac{7}{2}\right) = \frac{15}{8} \sqrt{\pi}$$ - $$\Gamma(5) = 4! = 24$$ 5. **Substitute values:** $$B\left(\frac{7}{2}, \frac{3}{2}\right) = \frac{\frac{15}{8} \sqrt{\pi} \times \frac{1}{2} \sqrt{\pi}}{24} = \frac{15}{8} \times \frac{1}{2} \times \frac{\pi}{24} = \frac{15 \pi}{384} = \frac{5 \pi}{128}.$$ 6. **Final integral value:** $$\int_0^{\frac{\pi}{2}} \cos^2 x \sin^6 x \, dx = \frac{1}{2} \times \frac{5 \pi}{128} = \frac{5 \pi}{256}.$$ **Note:** The user-provided answer is $$\frac{5 \pi}{32}$$, which is different. The correct evaluation using Beta function yields $$\frac{5 \pi}{256}$$. Please verify the problem statement or answer. **Summary:** $$\boxed{\int_0^{\frac{\pi}{2}} \cos^2 x \sin^6 x \, dx = \frac{5 \pi}{256}}.$$