# MIT Integration bee 2023 Regular Season $\int_0^\pi x\sin^4(x)dx$

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?

• Use $x\leftrightarrow\pi-x$ Commented Feb 16, 2023 at 0:04

## 2 Answers

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 $$I=\frac{\pi}{2}\int_0^\pi \sin^4x\mathrm{d}x=\frac{3\pi^2}{16}$$

• Jazak Allah for your answer! But I wonder what the formula for integrals with $\sin^{2n}(x)$ as the integrals. I forgot it unfortunately:( Commented Feb 16, 2023 at 2:28
• @KamalSaleh Salam, there are reduction formulas, but these "special" cases are known Commented Feb 16, 2023 at 22:35

Without using symmetry one can make use of the power reducing formula $$\sin ^4 x = \frac{3-4\cos2x+\cos4x}{8}$$ whence $$\int_{0}^{\pi}x\sin ^4 x \,dx =\frac{3}{8}\int_{0}^{\pi} x\,dx-\frac{1}{2}\int_{0}^\pi x\cos2x\, dx +\frac{1}{8}\int_{0}^{\pi}x\cos4x\, dx\tag{0}$$ so $$\int_{0}^{\pi}x\sin ^4 x \,dx =\frac{3}{16}\pi^2$$ since the the final two integrals in $$(0)$$ are equal to zero by an application of integration by parts.