微积分(A)随缘一题[25]

计算：\(\int_{-\pi}^{\pi} \frac{x\sin^3x}{1+e^x}dx\)

\[\int_{-\pi}^{\pi} \frac{x\sin^3x}{1+e^x}dx \\ =\int_{-\pi}^{0} \frac{x\sin^3x}{1+e^x}dx +\int_{0}^{\pi} \frac{x\sin^3x}{1+e^x}dx \\ =\int_{0}^{\pi} x\sin^3x \left(\frac{1}{1+e^{-x}}+\frac{1}{1+e^{x}}\right)dx \\ =\int_{0}^{\pi} x\sin^3x \left(\frac{e^{x}}{1+e^{x}}+\frac{1}{1+e^{x}}\right)dx \\ =\int_{0}^{\pi}x\sin^3xdx \\ =\int_{0}^{\pi}x\frac{3\sin x-\sin3x}{4}dx \\ =\frac{3}{4}\int_{0}^{\pi}x\sin x dx- \frac{1}{36}\int_{0}^{\pi}3x\sin3xd(3x) \\ =\frac{3}{4} (\sin x-x\cos x) \big|_{0}^{\pi}-\frac{1}{36}(\sin (3x)-3x\cos(3x)) \bigg|_{0}^{\pi} \\ =\frac{3}{4}[(0-\pi(-1))-(0-0)]-\frac{1}{36}[(0-3\pi(-1))-(0-0)] \\ =\frac{3\pi}{4}-\frac{\pi}{12}=\frac{2\pi}{3} \]