中册 4.1 不定积分计算 第3题

\section*{解题过程：} （1）方法 1：由于 $\displaystyle \int \frac{\mathrm{d} x}{1+\cos x}=\int \frac{\mathrm{d} x}{2 \cos ^{2} \frac{x}{2}}=\int \frac{1}{\cos ^{2} \frac{x}{2}} \mathrm{~d}\left(\frac{x}{2}\right)=\tan \frac{x}{2}+C$ ，所以 $$ \int \frac{\mathrm{d} x}{1+\sin x}=-\int \frac{1}{1+\cos \left(\frac{\pi}{2}-x\right)} \mathrm{d}\left(\frac{\pi}{2}-x\right)=-\tan \left[\frac{1}{2}\left(\frac{\pi}{2}-x\right)\right]+C=-\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)+C . $$ 方法 2： $\displaystyle \int \frac{\mathrm{d} x}{1+\sin x}=\int \frac{(1-\sin x) \mathrm{d} x}{1-\sin ^{2} x}=\int \frac{\mathrm{d} x}{\cos ^{2} x}+\int \frac{\mathrm{d} \cos x}{\cos ^{2} x}=\tan x-\frac{1}{\cos x}+C$ ． 方法3： $\displaystyle \int \frac{\mathrm{d} x}{1+\sin x}=\int \frac{\mathrm{d} x}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}}=\int \frac{\mathrm{d} x}{\cos ^{2} \frac{x}{2} \cdot\left(\tan \frac{x}{2}+1\right)^{2}}=2 \int \frac{\mathrm{~d} \tan \frac{x}{2}}{\left(\tan \frac{x}{2}+1\right)^{2}}=\frac{-2}{\tan \frac{x}{2}+1}+C$ ． 方法 4：令 $\displaystyle t=\tan \frac{x}{2}$ ，则 $\displaystyle \int \frac{\mathrm{d} x}{1+\sin x}=\int \frac{1}{1+\frac{2 t}{1+t^{2}}} \frac{2}{1+t^{2}} \mathrm{~d} t=2 \int \frac{1}{(t+1)^{2}} \mathrm{~d} t=\frac{-2}{t+1}+C=\frac{-2}{\tan \frac{x}{2}+1}+C$ ． （2）令 $\displaystyle t=\tan \frac{x}{2}$ ，则 $$ \begin{aligned} \int \frac{\mathrm{d} x}{1+4 \cos x} & =\int \frac{1}{1+4 \frac{1-t^{2}}{1+t^{2}}} \frac{2}{1+t^{2}} \mathrm{~d} t=2 \int \frac{1}{5-3 t^{2}} \mathrm{~d} t=-\frac{1}{\sqrt{5}} \int\left(\frac{1}{\sqrt{3} t-\sqrt{5}}-\frac{1}{\sqrt{3} t+\sqrt{5}}\right) \mathrm{d} t \\ & =\frac{1}{\sqrt{15}} \ln \frac{\sqrt{3} t+\sqrt{5}}{\sqrt{3} t-\sqrt{5}}+C=\frac{1}{\sqrt{15}} \ln \frac{\sqrt{3} \tan \frac{x}{2}+\sqrt{5}}{\sqrt{3} \tan \frac{x}{2}-\sqrt{5}}+C . \end{aligned} $$ （3） $\displaystyle \int \frac{\mathrm{d} x}{\cos x \sin ^{3} x}=\int \frac{\cos ^{2} x+\sin ^{2} x}{\cos x \sin ^{3} x} \mathrm{~d} x=\int \frac{\cos x}{\sin ^{3} x} \mathrm{~d} x+\int \frac{1}{\cos x \sin x} \mathrm{~d} x=\int \frac{\mathrm{d} \sin x}{\sin ^{3} x}+\int \csc 2 x \mathrm{~d}(2 x)$ $$ =-\frac{1}{2} \sin ^{2} x+\ln |\csc 2 x-\cot 2 x|+C . $$ （4）令 $\displaystyle t=\tan \frac{x}{2}$ ，则 $$ \int \frac{\mathrm{d} x}{2+\cos x+\sin x}=\int \frac{1}{2+\frac{1-t^{2}}{1+t^{2}}+\frac{2 t}{1+t^{2}}} \frac{2}{1+t^{2}} \mathrm{~d} t=2 \int \frac{1}{(t+1)^{2}+2} \mathrm{~d} t=\sqrt{2} \arctan \frac{t+1}{\sqrt{2}}+C $$ $$ =\sqrt{2} \arctan \left(\frac{1}{\sqrt{2}}\left(\tan \frac{x}{2}+1\right)\right)+C . $$ （5）设 $$ \int \frac{5 \sin x+2 \cos x}{\sin x+3 \cos x} \mathrm{~d} x=\int \frac{A(\sin x+3 \cos x)+B(\sin x+3 \cos x)^{\prime}}{\sin x+3 \cos x} \mathrm{~d} x=A x+B \ln |\sin x+3 \cos x|+C $$ 其中 $A, B$ 待定． 由 $5 \sin x+2 \cos x=A(\sin x+3 \cos x)+B(\cos x-3 \sin x)$ 得 $\displaystyle A=\frac{11}{10}, B=-\frac{13}{10}$ ．于是 $$ \int \frac{5 \sin x+2 \cos x}{\sin x+3 \cos x} \mathrm{~d} x=\int \frac{A(\sin x+3 \cos x)+B(\sin x+3 \cos x)^{\prime}}{\sin x+3 \cos x} \mathrm{~d} x=\frac{11}{10} x-\frac{13}{10} \ln |\sin x+3 \cos x|+C . $$ （6）令 $t=\sin ^{2} x$ ，则 $$ \begin{aligned} \int \frac{\ln \left(2+\sin ^{2} x\right)}{\left(1+\sin ^{2} x\right)^{2}} \sin 2 x \mathrm{~d} x & =\int \frac{\ln (2+t)}{(1+t)^{2}} \mathrm{~d} t=-\int \ln (2+t) \mathrm{d}\left(\frac{1}{1+t}\right)=-\frac{\ln (2+t)}{1+t}+\int \frac{1}{(1+t)(2+t)} \mathrm{d} t \\ & =-\frac{\ln (2+t)}{1+t}+\ln \frac{1+t}{2+t}+C=-\frac{\ln \left(2+\sin ^{2} x\right)}{1+\sin ^{2} x}+\ln \frac{1+\sin ^{2} x}{2+\sin ^{2} x}+C . \end{aligned} $$ （7）令 $t=\sqrt{1+\sin x}$ ，则 $$ \int \frac{\sqrt{1+\sin x}}{\cos x} \mathrm{~d} x=\int \frac{2 t^{2}}{2 t^{2}-t^{4}} \mathrm{~d} t=2 \int \frac{1}{2-t^{2}} \mathrm{~d} t=\frac{1}{\sqrt{2}} \ln \frac{|t+\sqrt{2}|}{|t-\sqrt{2}|}+C=\frac{1}{\sqrt{2}} \ln \frac{|\sqrt{1+\sin x}+\sqrt{2}|}{|\sqrt{1+\sin x}-\sqrt{2}|}+C $$