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Using the symmetry x\to\pi-x, evaluate I = \int_0^{\pi} x\,\sin^4…
Question
Using the symmetry x\to\pi-x, evaluate I = \int_0^{\pi} x\,\sin^4 x\,dx. It equals \frac{m}{n}\pi^2 with m/n in lowest terms. Find m+n.
✓ Verified answer: 19checked by our engine — not a guess
Step-by-step solution
Let I = \int_0^\pi x\sin^4 x\,dx. Substitute x\to\pi-x: I = \int_0^\pi (\pi-x)\sin^4 x\,dx = \pi\int_0^\pi \sin^4 x\,dx - I. Hence 2I = \pi\int_0^\pi \sin^4 x\,dx, so I = \frac{\pi}{2}\int_0^\pi \sin^4 x\,dx. Now \int_0^\pi \sin^4 x\,dx = 2\int_0^{\pi/2}\sin^4 x\,dx = 2\cdot\frac{3\cdot1}{4\cdot2}\cdot\frac{\pi}{2} = \frac{3\pi}{8}. Therefore I = \frac{\pi}{2}\cdot\frac{3\pi}{8} = \frac{3\pi^2}{16}. So m=3, n=16 and m+n = 19.
Final answer19
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