第四章总复习题课

一、选择题

1.设\begin{aligned} I=\int\frac{{\rm d}x}{({\rm e}^x+{\rm e}^{-x})} \end{aligned}，则\begin{aligned} I=[\ B\ ]. \end{aligned}

A.{\rm e}^x-{\rm e}^{-x}+C

B.{\rm arctan}\ {\rm e}^x+C

C.{\rm arctan}\ {\rm e}^{-x}+C

D.{\rm e}^x+{\rm e}^{-x}+C

答案：B.{\rm arctan}\ {\rm e}^x+C

解析： \begin{aligned} &u={\rm e}^x;\int \frac{1}{u^2+1}{\rm d}u={\rm arctan} \ u\\ \end{aligned}

\begin{aligned} I = \int \frac{{\rm d}x}{{\rm e}^x+{\rm e}^{-x}} &=\int\frac{{\rm e}^{x}}{({\rm e}^x)^2+1}{\rm d}x\\ &=\int \frac{{\rm d}({\rm e}^x)}{({\rm e}^{x})^2+1}\\ & ={\rm arctan}\ {\rm e}^x+C \end{aligned}

2.\begin{aligned} \int \frac1{\sqrt x {\rm cos}^{2}\sqrt x}{\rm d}x=[\ B\ ] \end{aligned}

A.\begin{aligned} \frac{1}{2} \ {\rm tan}\ \sqrt x+C \end{aligned}

B.\begin{aligned} 2{\rm tan}\ \sqrt x+C \end{aligned}

C.\begin{aligned} -\frac12{\rm tan}\ \sqrt x+C \end{aligned}

D.\begin{aligned} -2{\rm tan}\ \sqrt x+C \end{aligned}

答案：B.\begin{aligned} 2{\rm tan}\ \sqrt x+C \end{aligned}

解析：\begin{aligned} \int\frac 1{\sqrt x {\rm cos}^2\sqrt x}{\rm d}x &=2\int \frac1{2\sqrt x \cdot {\rm cos}^2\sqrt x}{\rm d}x \\ &=2\int\frac1{{\rm cos}^2\sqrt x}{\rm d}(\sqrt x)\\ &=2{\rm tan}\ \sqrt x+C \end{aligned}

3.\begin{aligned} \int x \cdot {\rm e}^{-x}{\rm d}x=[\ A\ ] \end{aligned}

A.-(x+1){\rm e}^{-x}+C

B.(x+1){\rm e}^{-x}+C

C.(x-1){\rm e}^{-x}+C

D.-(x-1){\rm e}^{-x}+C

答案：A.-(x+1){\rm e}^{-x}+C

解析：\begin{aligned} &u=x;v'={\rm e}^{-x}\\ &u'=1;v={\rm -e}^{-x}\\ \end{aligned}\\ \int u{\rm d}v=uv-\int v{\rm d}u \iff \int uv'{\rm d}x=uv-\int u'v{\rm d}x

\begin{aligned} \int x\cdot {\rm e}^{-x}{\rm d}x &=\int x\cdot {\rm d}({\rm -e}^{-x})\\ &=-x{\rm e}^{-x}-\int{\rm -e}^{-x}{\rm d}x\\ &=-x{\rm e}^{-x}+\int {\rm e}^{-x}{\rm d}x\\ &=-x{\rm e}^{-x}-{\rm e}^{-x}+C\\ &=-(x+1){\rm e}^{-x}+C \end{aligned}

4.\begin{aligned} \int x {\rm sin}\ x \ {\rm d}x=[\ A \ ] \end{aligned}

A.{\rm sin}\ x-x{\rm cos}\ x+C

B.{\rm sin}\ x+x{\rm cos}\ x+C

C.{\rm -sin}\ x+x{\rm cos}\ x+C

D.{\rm -sin}\ x-x{\rm cos}\ x+C

答案：A.{\rm sin}\ x-x{\rm cos}\ x+C

解析：\begin{aligned} &u=x;{\rm d}v={\rm d}(-{\rm cos}\ x)\\ &{\rm d}u={\rm d}x;v={\rm -cos}\ x \end{aligned}

\begin{aligned} \int x{\rm sin}\ x{\rm d}x &=\int x{\rm d}(-{\rm cos}\ x)\\ &=-x{\rm cos}\ x-\int-{\rm cos}\ x{\rm d}x\\ &=-x{\rm cos}\ x+\int {\rm cos}\ x\ {\rm d}x\\ &=-x{\rm cos}\ x+{\rm sin}\ x+C \end{aligned}

二、填空题

5.已知\begin{aligned} \int x f(x){\rm d}x={\rm cos}\ x+C.\ \ f(x)=[\ -\frac{{\rm sin}\ x}{x}\ ] \end{aligned}

解析：\begin{aligned} [\int f(x){\rm d}x]'=[F(x)+C]'=f(x)\\ \end{aligned}

\begin{aligned} xf(x)&=[\int xf(x){\rm d}x]'\\ &=({\rm cos}\ x+C)'\\ &=-{\rm sin}\ x\\ f(x)=- \frac{{\rm sin}\ x}{x}\\ \end{aligned}

6.\begin{aligned} \int f''(1+2x){\rm d}x=[\ \frac 12 f'(1+2x)+C\ ] \end{aligned}

解析：\begin{aligned} &\int F'(x){\rm d}x=\int f(x){\rm d}x=F(x)+C\\ &u=1+2x;{\rm d}u=2{\rm d}x;\int f''(u){\rm d}u \\ \end{aligned}

\begin{aligned} \int f''(1+2x){\rm d}x &=\frac12 \int f''(1+2x){\rm d}(1+2x)\\ &=\frac 12 f'(1+2x)+C \end{aligned}

三、计算题

7.计算不定积分\begin{aligned} \int (\frac {{\rm cos}\ \sqrt x}{\sqrt x}){\rm d}x \end{aligned}.

解答：\begin{aligned} &u=\sqrt x;{\rm d}u=\frac 1{2\sqrt x}{\rm d}x \\ &\int {\rm cos}\ u \ {\rm d}u={\rm sin}\ u\\ \end{aligned}

\begin{aligned} \int (\frac {{\rm cos}\ \sqrt x}{\sqrt x}){\rm d}x &=2\int {\rm cos}\ \sqrt x \cdot {\rm d}\sqrt x\\ &=2{\rm sin}\ \sqrt x+C \end{aligned}

8.计算不定积分\begin{aligned} \int {\rm cot}\ x\ {\rm d}x .\\ \end{aligned}

解答：\begin{aligned} &u={\rm sin}\ x;{\rm d}u={\rm cos}\ x{\rm d}x \\ &\int \frac 1{u}{\rm d}u=\ln|u|+C\\ \end{aligned}

\begin{aligned} \int {\rm cot}\ x{\rm d}x &=\int \frac {{\rm cos}\ x}{{\rm sin}\ x}{\rm d}x\\ &=\int \frac 1{{\rm sin}\ x}{\rm d}{\rm sin}\ x\\ &=\ln|{\rm sin}\ x|+C \end{aligned}

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