How do I find $$\int_0^\pi x\sin^4xdx$$? This is the 8th question in the MIT integration bee regular season. You could find integrals here. Integration by parts directly is out of question. I thought of substituting $1-\cos^2x=\sin^2x$, but that is just more complicated. Any ideas?
MIT Integration bee 2023 Regular Season $\int_0^\pi x\sin^4(x)dx$
calculuscontest-mathdefinite integralsintegration
Related Question
- Integral from MIT Integration Bee 2023 Quarterfinals – $\lim_{n\to\infty} \frac{1}{n} \int_{0}^{n} \cos^2(\pi x^2/\sqrt{2}) \, \mathrm{d}x$
- 2023 MIT Integration Bee Regular Season Problem 6
- Integral from MIT Integration Bee 2023 Semifinals – $\int_{0}^{\pi} \frac{2\cos(x)-\cos(2021x)-2\cos(2022x)-\cos(2023x)+2}{1-\cos(2x)}\,\textrm{d}x$
- Solve MIT Integration Bee $2023$: $\int_0^{100} \lfloor x\rfloor x\lceil x\rceil dx$
- Regular Season Problem 11 from 2023 MIT Integration Bee
Best Answer
Let $$I=\int_0^\pi x\sin^4x\,\mathrm{d}x$$ substituting $u=\pi-x$ you have: $$I=\int_0^\pi (\pi-x)\sin^4(\pi-x)\mathrm{d}x=\int_0^\pi (\pi-x)\sin^4x\mathrm{d}x=\pi\int_0^\pi \sin^4x\mathrm{d}x-I $$ Hence $$I=\frac{\pi}{2}\int_0^\pi \sin^4x\mathrm{d}x=\frac{3\pi^2}{16}$$