中册 4.2 定积分计算 第5题

\section*{解题过程：} 利用分部积分法计算． （1） $\displaystyle \int_{0}^{1} x^{2} \arctan x \mathrm{~d} x=\frac{1}{3} \int_{0}^{1} \arctan x \mathrm{~d}\left(x^{3}\right)=\frac{1}{3}\left(\left.x^{3} \arctan x\right|_{0} ^{1}-\int_{0}^{1} x^{3} \mathrm{~d}(\arctan x)\right)=\frac{1}{3}\left(\frac{\pi}{4}-\int_{0}^{1} \frac{x^{3}}{1+x^{2}} \mathrm{~d} x\right)$ $$ =\frac{\pi}{12}-\left.\frac{1}{3}\left(\frac{1}{2} x^{2}-\frac{1}{2} \ln \left(x^{2}+1\right)\right)\right|_{0} ^{1}=\frac{\pi-2+2 \ln 2}{12} . $$ （2） $\displaystyle \int_{0}^{\pi} x \sin ^{2} x \mathrm{~d} x=\frac{1}{2} \int_{0}^{\pi} x^{2}(1-\cos 2 x) \mathrm{d} x=\frac{1}{2} \int_{0}^{\pi} x^{2} \mathrm{~d} x-\frac{1}{4} \int_{0}^{\pi} x^{2} \mathrm{~d}(\sin 2 x)=\left.\frac{1}{6} x^{3}\right|_{0} ^{\pi}-\frac{1}{4} \int_{0}^{\pi} x^{2} \mathrm{~d} \sin 2 x$ $$ \begin{aligned} & =\frac{\pi^{3}}{6}-\left.\frac{1}{4} x^{2} \sin 2 x\right|_{0} ^{\pi}+\frac{1}{2} \int_{0}^{\pi} x \sin 2 x \mathrm{~d} x=\frac{\pi^{3}}{6}-\frac{1}{6} \int_{0}^{\pi} x \mathrm{~d}(\cos 2 x) \\ & =\frac{\pi^{3}}{6}-\left.\frac{1}{4} x \cos 2 x\right|_{0} ^{\pi}+\frac{1}{4} \int_{0}^{\pi} \cos 2 x \mathrm{~d} x=\frac{\pi^{3}}{6}-\frac{\pi}{4} \end{aligned} $$ （3） $\displaystyle \int_{0}^{2 \pi} x \cos ^{2} x \mathrm{~d} x=\frac{1}{2} \int_{0}^{2 \pi} x(1+\cos 2 x) \mathrm{d} x=\left.\frac{1}{4} x\right|_{0} ^{2 \pi}+\frac{1}{4} \int_{0}^{2 \pi} x \mathrm{~d}(\sin 2 x)$ $$ =\pi^{2}+\left.\frac{1}{4} x \sin 2 x\right|_{0} ^{2 \pi}-\frac{1}{4} \int_{0}^{2 \pi} \sin 2 x \mathrm{~d} x=\pi^{2} . $$ （4） $\int_{0}^{\pi} x^{2} \sqrt{1-\cos 2 x} \mathrm{~d} x=\sqrt{2} \int_{0}^{\pi} x^{2} \sin x \mathrm{~d} x=-\sqrt{2} \int_{0}^{\pi} x^{2} \mathrm{~d}(\cos x)=-\left.\sqrt{2} x^{2} \cos x\right|_{0} ^{\pi}+2 \sqrt{2} \int_{0}^{\pi} x \mathrm{~d}(\sin x)$ $$ =\sqrt{2} \pi^{2}+2 \sqrt{2}\left(\left.x \sin x\right|_{0} ^{\pi}-\int_{0}^{\pi} \sin x \mathrm{~d} x\right)=\sqrt{2} \pi^{2}+\left.2 \sqrt{2} \cos x\right|_{0} ^{\pi}=\sqrt{2} \pi^{2}-4 \sqrt{2} . $$ （5）令 $t=\arcsin x$ ，则 $x=\sin t, \mathrm{~d} x=\cos t \mathrm{~d} t$ 。于是 $$ \begin{aligned} \int_{0}^{\frac{1}{2}} \frac{\arcsin x}{\left(1-x^{2}\right)^{\frac{3}{2}}} \mathrm{~d} x & =\int_{0}^{\frac{\pi}{4}} \frac{t}{\cos ^{3} t} \cdot \cos t \mathrm{~d} t=\int_{0}^{\frac{\pi}{4}} t \sec ^{2} t \mathrm{~d} t=\int_{0}^{\frac{\pi}{4}} t \mathrm{~d}(\tan t)=\left.(t \tan t)\right|_{0} ^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \tan t \mathrm{~d} t \\ & =\frac{\pi}{4}-\int_{0}^{\frac{\pi}{4}} \frac{-\mathrm{d}(\cos t)}{\cos t}=\frac{\pi}{4}+\left.\ln \cos t\right|_{0} ^{\frac{\pi}{4}}=\frac{\pi}{4}+\ln \frac{\sqrt{2}}{2} . \end{aligned} $$ （6）令 $t=\sqrt{x}$ ，则 $$ \int_{0}^{1} \frac{\arcsin \sqrt{x}}{\sqrt{x(1-x)}} \mathrm{d} x=\int_{0}^{1} \frac{\arcsin t}{t \sqrt{1-t^{2}}} 2 t \mathrm{~d} t=2 \int_{0}^{1} \frac{\arcsin t}{\sqrt{1-t^{2}}} \mathrm{~d} t=2 \int_{0}^{1} \arcsin t \mathrm{~d}(\arcsin t)=\left.(\arcsin t)^{2}\right|_{0} ^{1}=\frac{\pi^{2}}{4} . $$