不定积分的计算

单变量函数的积分

张瑞
中国科学技术大学数学科学学院

三角函数

$\sin x, \cos x,\tan x,\cot x, \sec x, \csc x$

可以写成$R(\sin x, \cos x)$

对$I=\displaystyle\int R(\sin x, \cos x)dx$用万能公式$t=\tan\frac{x}2$，则

\[\begin{aligned} \sin x=\dfrac{2t}{1+t^2}, \cos x=\dfrac{1-t^2}{1+t^2}, \\ dx=d(2\arctan t)=\dfrac2{1+t^2}dt \end{aligned} \]

则

\[I=\int R(\dfrac{1-t^2}{1+t^2},\dfrac{2t}{1+t^2})\dfrac{2}{1+t^2}dt \]

例 1. $\displaystyle\int\dfrac1{2\sin x-\cos x+5}dx$

例 2. $\displaystyle\int\dfrac{\sin x\cos x}{\sin x+\cos x}dx$

$\displaystyle\int\dfrac1{2\sin x-\cos x+5}dx$

$t=\tan\dfrac{x}2$，则

\[\begin{aligned} \int\dfrac1{2\sin x-\cos x+5}dx =&\int\dfrac{\frac2{1+t^2}}{2\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}+5}dt \\ =&\int\dfrac2{4t-(1-t^2)+5(1+t^2)}dt \\ =&\int\dfrac1{3t^2+2t+2}dt \\ =&\dfrac13\int\dfrac1{(t+\frac13)^2+\frac59}dt \end{aligned} \]

$\displaystyle\int\dfrac{\sin x\cos x}{\sin x+\cos x}dx$

$t=\tan\dfrac{x}2$，则

\[=\int\dfrac{4t(1-t^2)}{(1+t^2)^2(-t^2+2t+1)}dt \]
\[\dfrac{4t(1-t^2)}{(1+t^2)^2(-t^2+2t+1)} =(\dfrac{t-1}{1+t^2})'+\dfrac1{t^2-2t-1} \]

$\displaystyle\int\dfrac{\sin x\cos x}{\sin x+\cos x}dx$

\[\begin{aligned} \int\dfrac{\sin x\cos x}{\sin x+\cos x}dx =&\int\dfrac{\sin x\cos x(-\sin x+\cos x)}{\cos^2x-\sin^2x}dx \\ =&\int\dfrac{\cos^2x\sin x}{2\cos^2x-1}dx+\int\dfrac{-\sin^2x\cos x}{1-2\sin^2x}dx \\ =&\int\dfrac{-\cos^2x}{2\cos^2x-1}d(\cos x)+\int\dfrac{-\sin^2x}{1-2\sin^2x}d(\sin x) \\ \end{aligned} \]
  • 若$R(-\sin x, \cos x)=-R(\sin x ,\cos x)$， 令$t=\cos x$;
  • 若$R(\sin x, -\cos x)=-R(\sin x ,\cos x)$， 令$t=\sin x$;
  • 若$R(-\sin x, -\cos x)=R(\sin x ,\cos x)$， 令$t=\tan x$;

如$P(\cos^2x, \cos x\sin x, \sin^2x)$, 令$t=\tan x$, 化为

\[\int P(\dfrac{1}{1+t^2},\dfrac{t}{1+t^2},\dfrac{t^2}{1+t^2})\dfrac1{1+t^2}dt \]

例 3. $\displaystyle\int\dfrac1{\sin^4x+\cos^4x}dx$

例 4. $\displaystyle\int\sin^2x\cos^3xdx$

例 5. $\displaystyle\int\dfrac{\sin^5x}{\cos^4x}dx$

例 6. $\displaystyle\int\dfrac1{\sin^4x\cos^2x}dx$

\[\int\dfrac1{\sin^4x+\cos^4x}dx=\int\dfrac{\sec^4x}{1+\tan^4x}dx \]

令$t=\tan x$，则有

\[\begin{aligned} =&\int\dfrac{1+t^2}{1+t^4}dt =\int\dfrac{1+\frac1{t^2}}{t^2+\frac1{t^2}}dt \\ =&\int\dfrac1{(t-\frac1t)^2+2}d(t-\frac1t) \end{aligned} \]

$\displaystyle\int\sin^2x\cos^3xdx$ 令$t=\sin x$，则

\[=\int t^2(1-t^2)dt=\dfrac{t^3}3-\dfrac{t^5}t+c \]

$\displaystyle\int\dfrac{\sin^5x}{\cos^4x}dx$ 令$t=\cos x$，则

\[=-\int\dfrac{t^4-2t^2+1}{t^4}dt=-t-\dfrac2t+\dfrac1{3t^3}+c \]

$\displaystyle\int\dfrac1{\sin^4x\cos^2x}dx$ 令$t=\tan x$，

\[=\int\dfrac{(1+t^2)^2}{t^4}dt=t-\dfrac2t-\dfrac1{3t^3}+c \]

例 7. $\displaystyle\int\dfrac{1}{A\cos^2x+B\sin x\cos x+C\sin^2x}dx$, $AC-B^2>0$, $x\in(\dfrac{\pi}2,\dfrac{\pi}2)$

例 8. $\displaystyle\int\dfrac{\sin^2 x\cos x}{\sin x+\cos x}dx$

例 9. $\displaystyle\int\dfrac1{\sin x\cos 2x}$

$\displaystyle\int\dfrac{1}{A\cos^2x+B\sin x\cos x+C\sin^2x}dx$

$t=\tan x$

\[\begin{aligned} =&\int\dfrac1{A+2Bt+Ct^2}dt \\ =&\dfrac1{\sqrt{AC-B^2}}\arctan\dfrac{C\tan x+B}{\sqrt{AC-B^2}}+c \end{aligned} \]

$\displaystyle\int\dfrac{\sin^2 x\cos x}{\sin x+\cos x}dx$

$t=\tan x$

$\displaystyle\int\dfrac1{\sin x\cos 2x}=\int\dfrac1{\sin x(2\cos^2x-1)}dx$

$t=\cos x$

\[\begin{aligned} =&\int\dfrac1{(1-t^2)(1-2t^2)}dt \\ =&\dfrac1{\sqrt 2}\ln|\dfrac{1+t\sqrt 2}{1-t\sqrt 2}|+\ln|\dfrac{1-t}{1+t}|+c \end{aligned} \]

若$R(\sin(x),\cos(x))$不满足上述的三个条件，则可以对函数$R(\cdot,\cdot)$做如下分解

\[\begin{aligned} R(u,v)=&\frac{R(u,v)-R(-u,v)}2+\frac{R(-u,v)-R(-u,-v)}2 \\ &+\frac{R(-u,-v)+R(u,v)}2 \end{aligned} \]

分解出的三个部分, 分别满足三个条件

$\displaystyle\int\sin^px\cos^qxdx$ 形态

例 10. $\displaystyle\int\dfrac1{\cos^5x}dx$

例 11. $\displaystyle\int {\cos^5x}dx$

例 12. $\displaystyle\int \frac{\cos^4x}{\sin^3x}dx$

\[\int\dfrac1{\cos^5x}dx=\dfrac{\sin x}{4\cos^4x}+\dfrac34\int\dfrac1{\cos^3x}dx \]
\[\int\dfrac1{\cos^3x}dx=\dfrac{\sin x}{2\cos^2x}+\dfrac12\int\dfrac1{\cos x}dx \]

$\displaystyle\int\sin^px\cos^qxdx=\int\frac12\sin^{p-1}x(1-\sin^2x)^{\frac{p-1}2}2\sin x\cos xdx$

令$z=\sin^2x$，则$dz=2\sin x\cos xdx$，有

\[\int\sin^px\cos^qxdx=\frac12\int(1-z)^{\frac{p-1}2}z^{\frac{q-1}2}dz \]

利用前面的结论，可以得到如下递推关系

  • $q\neq-1$ ,
    \[=-\dfrac{\sin^{p+1}x\cos^{q+1}x}{q+1}+\dfrac{p+q+2}{q+1}\displaystyle\int\sin^px\cos^{q+2}xdx \]
  • $p\neq-1$,
    \[=\dfrac{\sin^{p+1}x\cos^{q+1}x}{p+1}+\dfrac{p+q+2}{p+1}\displaystyle\int\sin^{p+2}x\cos^{q}xdx \]
  • $p+q\neq0$,
    \[\begin{aligned} = \dfrac{\sin^{p+1}x\cos^{q-1}x}{p+q}+\dfrac{q-1}{p+q}\displaystyle\int\sin^{p}x\cos^{q-2}xdx \\ = -\dfrac{\sin^{p-1}x\cos^{q+1}x}{p+q}+\dfrac{p-1}{p+q}\displaystyle\int\sin^{p-2}x\cos^{q}xdx \end{aligned} \]

当$p,q$为整数，可以得到递推公式。最终化为以下积分

\[\begin{aligned} &\int \cos xdx=\sin x &,& \int \sin xdx=-\cos x \\ &\int \dfrac1{\cos x}dx=\ln\left|\tan(\frac{x}2+\dfrac{\pi}4)\right| &,& \int \dfrac1{\sin x}dx=\ln\left|\tan(\frac{x}2)\right|, \\ &\int \sin x\cos xdx=\dfrac{\sin^2x}2 &, & \int\dfrac{1}{\sin x\cos x}dx=\ln|\tan x| \\ &\int \dfrac{\sin x}{\cos x}dx=-\ln|\cos x| &, & \int \dfrac{\cos x}{\sin x}dx=\ln|\sin x| , \end{aligned} \]

例 13. $\displaystyle\int\dfrac{\sin x}{a\cos x+b\sin x}dx$,$\displaystyle\int\dfrac{\cos x}{a\cos x+b\sin x}dx$,

例 14. $\displaystyle\int\dfrac{\sin x}{a\cos x+b\sin x+c}dx$,$\displaystyle\int\dfrac{\cos x}{a\cos x+b\sin x+c}dx$,

例 15. $\displaystyle\int\dfrac{1}{a+b\cos x}dx$,$\displaystyle\int\dfrac{1}{a+b\sin x}dx$,

例 16. $\displaystyle\int\dfrac{1}{a+b\cos x+c\sin x}dx$,

$T_1=\displaystyle\int\dfrac{\sin x}{a\cos x+b\sin x}dx$,$T_2=\displaystyle\int\dfrac{\cos x}{a\cos x+b\sin x}dx$,

\[\begin{aligned} aT_1+bT_2&=\int 1 dx=x \\ -aT_1+bT_2&=\int\dfrac{-a\sin x+b\cos x}{a\cos x+b\sin x}dx \\ &=\int\dfrac1{a\cos x+b\sin x}d{(a\cos x+b\sin x)} \end{aligned} \]
\[\begin{aligned} T_1=\dfrac{1}{a^2+b^2}(bx-a\ln|a\cos x+b\sin x|)+c \\ T_2=\dfrac{1}{a^2+b^2}(ax+b\ln|a\cos x+b\sin x|)+c \\ \end{aligned} \]

$\displaystyle\int\dfrac{1}{a+b\cos x}dx$， 令 $t=\tan(\frac{x}2)$

\[=\int\dfrac{2}{(a+b)+(a-b)t^2}dt \]
  • $a>b,a>0,b>0$或$a<b,a<0,b<0$时，即$|a|>|b|$，有$=\dfrac{2}{\sqrt{a^2-b^2}}\arctan(\sqrt{\dfrac{a-b}{a+b}}t)+c$
  • $|a|<|b|$，有$=\dfrac{1}{\sqrt{b^2-a^2}}\ln|\dfrac{\sqrt{b+a}+t\sqrt{b-a}}{\sqrt{b+a}-t\sqrt{b-a}}|+c$

$\displaystyle\int\dfrac{1}{a+b\sin x}dx=\displaystyle\int\dfrac{1}{a+b\cos(\dfrac{\pi}2-x}dx$

$\displaystyle\int\dfrac{1}{a+b\cos x+c\sin x}dx=\displaystyle\int\dfrac{1}{a+\sqrt{b^2+c^2}\cos(x-\alpha)}dx$

其中

\[\cos\alpha=\dfrac{b}{\sqrt{b^2+c^2}}, \sin\alpha=\dfrac{c}{\sqrt{b^2+c^2}}, \]
\[\int\dfrac{a_1\sin x+b_1\cos x}{a \sin x+b \cos x}dx=A x+B\ln|a\sin x+b\cos x|+c \]
\[\begin{aligned} \int\dfrac{a_1\sin x+b_1\cos x+c_1}{a \sin x+b \cos x+c}dx=A x+B\ln|a\sin x+b\cos x+c| \\ +C\int\dfrac{1}{a\sin x+b\cos x+c}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac{a_1\sin^2 x+2b_1\sin x\cos x+c_1\cos^2 x}{a \sin x+b \cos x}dx=A\sin x+B\cos x \\ +C\int\dfrac1{a\sin x+b\cos x}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac1{(a\sin x+b\cos x)^n}dx=\dfrac{A\sin x+B\cos x}{(a\sin x+b\cos x)^{n-1}} \\ +C\int\dfrac1{(a\sin x+b\cos x)^{n-2}}dx \end{aligned} \]
\[\begin{aligned} \int\dfrac1{(a+b\cos x)^n}dx=\dfrac{A\sin x}{(a+b\cos x)^{n-1}}+B\int\dfrac1{(a+b\cos x)^{n-1}}dx \\ +C\int\dfrac{1}{(a+b\cos x)^{n-2}}dx \end{aligned} \]

例 17. $\displaystyle\int\dfrac{\sin(2x)}{\sqrt{2-\sin^4x}}dx$

例 18. $\displaystyle\int\cos^5x\sqrt{\sin x}dx$

三角函数化

例 19. $\displaystyle\int\dfrac1{(1-x^2)^{\frac32}}dx$

例 20. $\displaystyle\int\dfrac1{(1+x^2)\sqrt{1-x^2}}dx$

$\displaystyle\int\cos^5x\sqrt{\sin x}dx$

令$t=1-5x^2$，则$-10xdx=dt$

\[\begin{aligned} \int\cos^5x\sqrt{\sin x}dx=\int x^2x(1-5x^2)^{10}\dfrac1{-10x}dt \\ =\int\dfrac{1-t}5t^{10}\dfrac1{-10}dt \end{aligned} \]

$\displaystyle\int\dfrac{\sin^2x}{\cos^6x}dx$

令$t=\tan x$，则有

\[\begin{aligned} =\int\tan^2x(1+\tan^2x)d(\tan x) \\ =\int t^2(1+t^2)dt \end{aligned} \]

$\displaystyle\int\dfrac1{(1-x^2)^{\frac32}}dx$

$x=\sin t$， $dx=\cos t dt$，

\[\begin{aligned} \int\dfrac1{(1-x^2)^{\frac32}}dx=\int\dfrac{\cos t}{\cos^3 t}dt \\ =\int\dfrac1{\cos^2t}dt=\tan t+c \\ =\dfrac{x}{\sqrt{1-x^2}}+c \end{aligned} \]

$\displaystyle\int\dfrac1{(1-x^2)^{\frac32}}dx$

\[\begin{aligned} \int\dfrac1{(1-x^2)^{\frac32}}dx=\int\dfrac{1}{x^3(\frac1{x^2}-1)^{\frac32}}dt \\ =-\dfrac12\int\dfrac{1}{(\frac1{x^2}-1)^{\frac32}}d(\dfrac1{x^2}) =-\dfrac12\int\dfrac1{(\frac1{x^2}-1)^{\frac32}}d(\dfrac1{x^2}-1) \\ =(\dfrac1{x^2}-1)^{-\frac12}+c \end{aligned} \]
\[\begin{aligned} \int\dfrac{\sin(2x)}{\sqrt{2-\sin^4x}}dx =\int\dfrac{2\sin x\cos x}{\sqrt{2-\sin^4x}}dx \\ \end{aligned} \]
\[=\int\dfrac{1}{\sqrt{2-\sin^4x}}d(\sin^2 x) \]
\[\begin{aligned} =\int\dfrac{\sqrt 2}{\sqrt 2\sqrt{1-(\frac{\sin^2x}{\sqrt 2})^2}}d(\frac{\sin^2 x}{\sqrt 2}) \\ \end{aligned} \]
\[=\int\dfrac{1}{\sqrt{1-t^2}}dt=\arcsin\dfrac{\sin^2x}{\sqrt 2}+c \]

杂题

例 21. $\displaystyle\int\dfrac{e^x\cos x-e^x\sin x}{e^{2x}}dx$

例 22. $\displaystyle\int\dfrac{x\cos x-\sin x}{x^2}dx$

例 23. $\displaystyle\int\dfrac{x\cos x-\sin x}{x^2\sqrt{1-\frac{\sin^2x}{x^2}}}dx$

\[\begin{aligned} \int\dfrac{e^x\cos x-e^x\sin x}{e^{2x}}dx=\int d\dfrac{\sin x}{e^x} \\ =\dfrac{\sin x}{e^x}+c \end{aligned} \]
\[\int\dfrac{x\cos x-\sin x}{x^2}dx=\int d\dfrac{\sin x}{x} \]
\[\begin{aligned} \int\dfrac{x\cos x-\sin x}{x^2\sqrt{1-\frac{\sin^2x}{x^2}}}dx =\int\dfrac1{\sqrt{1-(\frac{\sin x}{x})^2}}d\dfrac{\sin x}{x} \\ =\arcsin\dfrac{\sin x}x+c \end{aligned} \]

例 24. $\displaystyle\int x(1-x)^{10}dx$

例 25. $\displaystyle\int\dfrac{x^2}{(1-x)^{100}}dx$

例 26. $\displaystyle\int x^3(1-5x^2)^{10}dx$

例 27. $\displaystyle\int x\sqrt{2-5x}dx$

\[\begin{aligned} \int x(1-x)^{10}dx=\int (-(1-x)+1)(1-x)^{10}dx \\ =-\int(1-x)dx+\int(1-x)^{10}dx \end{aligned} \]

令$t=(1-x)$，则

\[\begin{aligned} \int x(1-x)^{10}dx=\int (1-t)t^{10}dx \\ =-\int t^{10}dt+\int t^{11}dx \end{aligned} \]

$\displaystyle\int\dfrac{x^2}{(1-x)^{100}}dx$

令$t=x-1$，则

\[\int\dfrac{x^2}{(1-x)^{100}}dx=\int\dfrac{(t+1)^2}{t^{100}}dt \]

$\displaystyle\int x^3(1-5x^2)^{10}dx$

$\displaystyle\int x\sqrt{2-5x}dx$

$t=2-5x$

本节读完

例 28. Thanks

28.