中册 4.2 定积分计算 第11题

\section*{解题过程：} （1） $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} \mathrm{~d} x=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\sin ^{4} x} \mathrm{~d}\left(\sin ^{2} x\right)=\left.\frac{1}{2} \arctan \left(\sin ^{2} x\right)\right|_{0} ^{\frac{\pi}{2}}=\frac{\pi}{8}$ ． （2） $\displaystyle \int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)^{2} d x=\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2}\left(x-\frac{\pi}{4}\right)}{\sin ^{2}\left(x+\frac{\pi}{4}\right)} d x=\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2}\left(x-\frac{\pi}{4}\right)}{\cos ^{2}\left(x-\frac{\pi}{4}\right)} d x=\int_{0}^{\frac{\pi}{4}}\left(\frac{1}{\cos ^{2}\left(x-\frac{\pi}{4}\right)}-1\right) d x$ $$ =\left.\tan \left(x-\frac{\pi}{4}\right)\right|_{0} ^{\frac{\pi}{4}}-\frac{\pi}{4}=-1-\frac{\pi}{4} . $$ （3） $\displaystyle \int_{0}^{\pi} \frac{3+\sin 2 x}{2+\cos x} \mathrm{~d} x=3 \int_{0}^{\pi} \frac{1}{2+\cos x} \mathrm{~d} x+2 \int_{0}^{\pi} \frac{\sin x \cos x}{2+\cos x} \mathrm{~d} x$ ． 由于 $\displaystyle \int_{0}^{\pi} \frac{\mathrm{d} x}{2+\cos x}=\int_{0}^{\pi} \frac{\mathrm{d} x}{2-\cos x}=\frac{1}{2} \int_{0}^{\pi} \frac{4 \mathrm{~d} x}{4-\cos ^{2} x}=4 \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d} x}{4-\cos ^{2} x}=4 \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d}(\tan x)}{4 \tan ^{2} x+3}=\frac{2}{\sqrt{3}} \frac{\pi}{2}=\frac{\pi}{\sqrt{3}}$ ， $$ \begin{aligned} \int_{0}^{\pi} \frac{\sin x \cos x}{2+\cos x} \mathrm{~d} x & =-\int_{0}^{\pi} \frac{\cos x}{2+\cos x} \mathrm{~d}(\cos x)=-\int_{0}^{\pi}\left(1-\frac{2}{2+\cos x}\right) \mathrm{d}(\cos x) \\ & =-\left.(\cos x-2 \ln (2+\cos x))\right|_{0} ^{\pi}=2-2 \ln 3 \end{aligned} $$ 所以 $\displaystyle \int_{0}^{\pi} \frac{3+\sin 2 x}{2+\cos x} \mathrm{~d} x=3 \int_{0}^{\pi} \frac{1}{2+\cos x} \mathrm{~d} x+2 \int_{0}^{\pi} \frac{\sin x \cos x}{2+\cos x} \mathrm{~d} x=\sqrt{3} \pi+4-4 \ln 3$ ． （4）方法 1： $\displaystyle \int_{0}^{2 \pi} \frac{\mathrm{~d} \theta}{2+\cos \theta}=\int_{-\pi}^{\pi} \frac{\mathrm{d} \theta}{2+\cos \theta}=2 \int_{0}^{\pi} \frac{\mathrm{d} \theta}{2+\cos \theta}$ ． 因为 $\displaystyle \int_{0}^{\pi} \frac{\mathrm{d} \theta}{2+\cos \theta}=\int_{0}^{\pi} \frac{\mathrm{d} \theta}{2-\cos \theta}=\frac{1}{2} \int_{0}^{\pi} \frac{4 \mathrm{~d} \theta}{4-\cos ^{2} \theta}=4 \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d} \theta}{4-\cos ^{2} \theta}=4 \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d}(\tan \theta)}{4 \tan ^{2} \theta+3}=\frac{\pi}{\sqrt{3}}$ ，故 $\displaystyle \int_{0}^{2 \pi} \frac{\mathrm{~d} \theta}{2+\cos \theta}=\frac{2 \sqrt{3}}{3} \pi$ ． 方法 2：记 $\displaystyle I=\int_{0}^{2 \pi} \frac{\mathrm{~d} \theta}{2+\cos \theta}=\int_{0}^{\pi} \frac{\mathrm{d} \theta}{2+\cos \theta}+\int_{\pi}^{2 \pi} \frac{\mathrm{~d} \theta}{2+\cos \theta}=\int_{0}^{\pi} \frac{\mathrm{d} \theta}{2+\cos \theta}+\int_{0}^{\pi} \frac{\mathrm{d} \theta}{2-\cos \theta}$ $$ =\int_{0}^{\pi}\left(\frac{1}{2+\cos \theta}+\frac{1}{2-\cos \theta}\right) \mathrm{d} \theta=4 \int_{0}^{\pi} \frac{1}{4-\cos ^{2} \theta} \mathrm{~d} \theta=8 \int_{0}^{\frac{\pi}{2}} \frac{1}{4-\cos ^{2} \theta} \mathrm{~d} \theta $$ 令 $\tan \theta=t$ ，则 $\displaystyle I=8 \int_{0}^{+\infty} \frac{1}{4-\frac{1}{1+t^{2}}} \frac{\mathrm{~d} t}{1+t^{2}}=8 \int_{0}^{+\infty} \frac{\mathrm{d} t}{4 t^{2}+3}=\frac{4}{\sqrt{3}} \cdot \frac{\pi}{2}=\frac{2}{\sqrt{3}} \pi$ ． （5） $\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{x}{1+\cos 2 x} \mathrm{~d} x=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{x}{\cos ^{2} x} \mathrm{~d} x=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} x \mathrm{~d}(\tan x)=\left.\frac{1}{2}(x \tan x+\ln \cos x)\right|_{0} ^{\frac{\pi}{4}}=\frac{\pi}{8}+\frac{1}{2} \ln \frac{\sqrt{2}}{2}$ ． （6） $\displaystyle \int_{0}^{\pi} \frac{\cos 4 \theta d \theta}{1+\cos ^{2} \theta}=\int_{0}^{\pi} \frac{\cos 4 \theta}{1+\frac{1+\cos 2 \theta}{2}} d \theta=2 \int_{0}^{\pi} \frac{\cos 4 \theta}{3+\cos 2 \theta} d \theta \xlongequal{t=2 \theta} \int_{0}^{2 \pi} \frac{\cos 2 t}{3+\cos t} d t$ $$ \begin{aligned} & =2 \int_{0}^{\pi} \frac{\cos 2 t}{3+\cos t} \mathrm{~d} t=2 \int_{0}^{\pi} \frac{2 \cos ^{2} t-1}{3+\cos t} \mathrm{~d} t=2 \int_{0}^{\pi}\left[-2(3-\cos t)+\frac{17}{3+\cos t}\right] \mathrm{d} t \\ & =-12 \pi+34 \int_{0}^{\pi} \frac{1}{3+\cos t} \mathrm{~d} t \xlongequal{u=\tan \frac{t}{2}}-12 \pi+34 \int_{0}^{+\infty} \frac{1}{2+u^{2}} \mathrm{~d} u=\left(\frac{17}{\sqrt{2}}-12\right) \pi \end{aligned} $$ （7）令 $\displaystyle \tan \frac{1}{2} \theta=t$ ，则 $$ \begin{aligned} \int_{0}^{\pi} \frac{\mathrm{d} \theta}{1-\sqrt{2} \cos \theta+a} & =2 \int_{0}^{+\infty} \frac{\mathrm{d} t}{(1+a-\sqrt{2})+(1+a+\sqrt{2}) t^{2}}=\frac{2}{(1+a-\sqrt{2})} \int_{0}^{+\infty} \frac{\mathrm{d} t}{1+(b t)^{2}} \\ & =\left.\frac{2}{b(1+a-\sqrt{2})} \arctan (b t)\right|_{0} ^{+\infty}=\frac{\pi}{b(1+a-\sqrt{2})}=\pi \sqrt{a^{2}+2 a-1} \end{aligned} $$ 其中 $\displaystyle b=\sqrt{\frac{1+a+\sqrt{2}}{1+a-\sqrt{2}}}$ ． （8） $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d} x}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x}=\frac{1}{|a b|} \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d}\left(\left|a b^{-1}\right| \tan x\right)}{\left(\left|a b^{-1}\right| \tan x\right)^{2}+1}=\left.\frac{1}{|a b|} \arctan \left(\left|a b^{-1}\right| \tan x\right)\right|_{0} ^{\frac{\pi}{2}}=\frac{\pi}{2|a b|}$ ．