中册 4.1 不定积分计算 第9题

\section*{解题过程：} （1）记 $I_{1}=\int \mathrm{e}^{a x} \cos b x \mathrm{~d} x, I_{2}=\int \mathrm{e}^{a x} \sin b x \mathrm{~d} x$ ，由分部积分法得 $$ \begin{aligned} & I_{1}=\frac{1}{a} \int \cos b x \mathrm{~d}\left(\mathrm{e}^{a x}\right)=\frac{1}{a}\left(\mathrm{e}^{a x} \cos b x+b \int \mathrm{e}^{a x} \sin b \mathrm{~d} x\right)=\frac{1}{a}\left(\mathrm{e}^{a x} \cos b x+b I_{2}\right), \\ & I_{2}=\frac{1}{a} \int \sin b x \mathrm{~d}\left(\mathrm{e}^{a x}\right)=\frac{1}{a}\left(\mathrm{e}^{a x} \sin b x-b I_{1}\right) . \end{aligned} $$ 解得 $\displaystyle I_{1}=\int \mathrm{e}^{a x} \cos b x \mathrm{~d} x=\mathrm{e}^{a x} \frac{b \sin b x+a \cos b x}{a^{2}+b^{2}}+C, I_{2}=\int \mathrm{e}^{a x} \sin b x \mathrm{~d} x=\mathrm{e}^{a x} \frac{a \sin b x-b \cos b x}{a^{2}+b^{2}}+C$ ． （2）记 $I=\int \cos (\ln x) \mathrm{d} x, J=\int \sin (\ln x) \mathrm{d} x$ ，由分部积分法得 $$ I=x \cos (\ln x)+\int \sin (\ln x) \mathrm{d} x=x \cos (\ln x)+J, J=x \sin (\ln x)-\int \cos (\ln x) \mathrm{d} x=x \sin (\ln x)-I $$ 解得 $\displaystyle I=\int \cos (\ln x) \mathrm{d} x=\frac{1}{2} x(\cos (\ln x)+\sin (\ln x))+C, J=\int \sin (\ln x) \mathrm{d} x=\frac{1}{2} x(\sin (\ln x)-\cos (\ln x))+C$ ． 注：令 $t=\ln x$ ，则 $\mathrm{d} x=\mathrm{e}^{t} \mathrm{~d} t, I=\int \mathrm{e}^{t} \cos t \mathrm{~d} t, J=\int \mathrm{e}^{t} \sin t \mathrm{~d} t$ ，本质上与 $(1)$ 是同一种类型题． （3） $\displaystyle \int \mathrm{e}^{x} \sin ^{2} x \mathrm{~d} x=\frac{1}{2} \int \mathrm{e}^{x} \mathrm{~d} x-\frac{1}{2} \int \mathrm{e}^{x} \cos 2 x \mathrm{~d} x=\frac{1}{2} \mathrm{e}^{x}-\frac{1}{2} \int \mathrm{e}^{x} \cos 2 x \mathrm{~d} x$ ． $$ \int \mathrm{e}^{x} \cos 2 x \mathrm{~d} x=\mathrm{e}^{x} \cos 2 x+2 \int \mathrm{e}^{x} \sin 2 x \mathrm{~d} x=\mathrm{e}^{x}(\cos 2 x+2 \sin 2 x)-4 \int \mathrm{e}^{x} \cos 2 x \mathrm{~d} x $$ 从而 $\displaystyle \int \mathrm{e}^{x} \cos 2 x \mathrm{~d} x=\frac{1}{5} \mathrm{e}^{x}(\cos 2 x+2 \sin 2 x)+C$ ， 于是 $\displaystyle \int \mathrm{e}^{x} \sin ^{2} x \mathrm{~d} x=\frac{1}{2} \mathrm{e}^{x}-\frac{1}{10} \mathrm{e}^{x}(\cos 2 x+2 \sin 2 x)+C$ ． （4） $\int x \mathrm{e}^{x} \cos x \mathrm{~d} x=\int x \cos x \mathrm{de}^{x}=x \mathrm{e}^{x} \cos x-\int \mathrm{e}^{x}(\cos x-x \sin x) \mathrm{d} x$ $$ \begin{aligned} & =x \mathrm{e}^{x} \cos x-\int \mathrm{e}^{x} \cos x \mathrm{~d} x+\int x \mathrm{e}^{x} \sin x \mathrm{~d} x \\ \int x \mathrm{e}^{x} \sin x \mathrm{~d} x & =\int x \sin x \mathrm{de}^{x}=x \mathrm{e}^{x} \sin x-\int \mathrm{e}^{x}(\sin x+x \cos x) \mathrm{d} x \\ & =x \mathrm{e}^{x} \sin x-\int \mathrm{e}^{x} \sin x \mathrm{~d} x-\int x \mathrm{e}^{x} \cos x \mathrm{~d} x \end{aligned} $$ 由 $\displaystyle \int \cos x \mathrm{e}^{x} \mathrm{~d} x=\frac{\mathrm{e}^{x}(\sin x+\cos x)}{2}, \int \sin x \mathrm{e}^{x} \mathrm{~d} x=\frac{\mathrm{e}^{x}(\sin x-\cos x)}{2}$ 得 $$ \begin{aligned} & \int x \mathrm{e}^{x} \cos x \mathrm{~d} x=\frac{x \mathrm{e}^{x}(\sin x+\cos x)}{2}-\frac{\mathrm{e}^{x} \cos x}{2}+C \\ & \int x \mathrm{e}^{x} \sin x \mathrm{~d} x=-\frac{x \mathrm{e}^{x}(\sin x-\cos x)}{2}+\frac{\mathrm{e}^{x} \sin x}{2}+C \end{aligned} $$ （5）令 $u=\arctan x$ ，则 $\displaystyle \mathrm{d} u=\frac{1}{1+x^{2}} \mathrm{~d} x$ ，则 $$ \int \frac{x \mathrm{e}^{\arctan x}}{\left(1+x^{2}\right)^{\frac{3}{2}}} \mathrm{~d} x=\int \frac{\tan u \mathrm{e}^{u}}{\sec u} \mathrm{~d} u=\int \mathrm{e}^{u} \sin u \mathrm{~d} u=\frac{1}{2} \mathrm{e}^{u}(\sin u-\cos u)+C=\frac{x-1}{2 \sqrt{1+x^{2}}} \mathrm{e}^{\arctan x}+C $$ （6） $\displaystyle \int \mathrm{e}^{\sin x} \frac{x \cos ^{3} x-\sin x}{\cos ^{2} x} \mathrm{~d} x=\int x \mathrm{de}^{\sin x}-\int \mathrm{e}^{\sin x} \mathrm{~d}\left(\frac{1}{\cos x}\right)=\mathrm{e}^{\sin x}\left(x-\frac{1}{\cos x}\right)+C$ ． （7） $\displaystyle \int \frac{\ln \sin x}{\sin ^{2} x} \mathrm{~d} x=-\int \ln \sin x \mathrm{~d}(\cot x)=-\cot x \ln \sin x+\int \cot x \frac{\cos x}{\sin x} \mathrm{~d} x$ $$ =-\cot x \ln \sin x+\int\left(\csc ^{2} x-1\right) \mathrm{d} x=-\cot x \ln \sin x-\cot ^{2} x-x+C . $$