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Using the Beta function (or Wallis reduction), evaluate I =…

Question

Using the Beta function (or Wallis reduction), evaluate I = \int_0^{\pi/2} \sin^6 x\,\cos^4 x\,dx. It equals \frac{m}{n}\pi with m/n in lowest terms. Find m+n.

✓ Verified answer: 515checked by our engine — not a guess

Step-by-step solution

Use \int_0^{\pi/2}\sin^{2a-1}x\cos^{2b-1}x\,dx = \frac12 B(a,b) = \frac{\Gamma(a)\Gamma(b)}{2\Gamma(a+b)}. Here 2a-1=6\Rightarrow a=7/2, 2b-1=4\Rightarrow b=5/2, a+b=6. So I = \frac{\Gamma(7/2)\Gamma(5/2)}{2\Gamma(6)}. With \Gamma(7/2)=\frac{15\sqrt\pi}{8}, \Gamma(5/2)=\frac{3\sqrt\pi}{4}, \Gamma(6)=120: numerator = \frac{15\sqrt\pi}{8}\cdot\frac{3\sqrt\pi}{4} = \frac{45\pi}{32}; divide by 2\cdot120=240: I = \frac{45\pi}{32\cdot240} = \frac{45\pi}{7680} = \frac{3\pi}{512}. So m=3, n=512 and m+n = 515.

Final answer515

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