不定积分基本公式与推导

1

$\int\ a^x\,dx=\dfrac{a^x}{lna} + C,\quad ({a^x})’ = a^x lna$

2

$\int\tan x\,dx=\int\dfrac{sinx}{cosx}\,dx=-\int\dfrac{1}{cosx}\,d(cosx)=-\ln{|\cos{x}|} + C$

$\int\cot x\,dx=\int\dfrac{cosx}{sinx}\,dx=-\int\dfrac{1}{sinx}\,d(sinx)=\ln{|\sin{x}|} + C$

3

$\int\dfrac{1}{cosx}\,dx =\int\sec{x}\,dx\\ =\int\dfrac{cosx}{(cosx)^2}\,dx =\int\dfrac{cosx}{1-(sinx)^2}\,dx \\ =\int\dfrac{1}{1-(sinx)^2}\,d(sinx)=\int\dfrac{1}{(1+sinx)(1-sinx)}\,d(sinx) \\ =\dfrac{1}{2}[\int\dfrac{1}{1+sinx}\,d(sinx) – \int\dfrac{1}{1-sinx}\,d(sinx)] \\ =\dfrac{1}{2}ln|\dfrac{1+sinx}{1-sinx}| + C \\ =\dfrac{1}{2}ln|\dfrac{(1+sinx)^2}{(cosx)^2}|=ln|\dfrac{1+sinx}{cosx}| \\ =ln|secx + tanx| + C$

$\int\dfrac{1}{sinx}\,dx =\int\csc{x}\,dx \\ =\int\dfrac{sinx}{(sinx)^2}\,dx =\int\dfrac{sinx}{1-(cosx)^2}\,dx \\ =-\int\dfrac{1}{1-(cosx)^2}\,d(cosx)=-\int\dfrac{1}{(1+cosx)(1-cosx)}\,d(cosx) \\ =-\dfrac{1}{2}[\int\dfrac{1}{1+cosx}\,d(cosx+1) – \int\dfrac{1}{1-cosx}\,d(1-cosx)] \\ =-\dfrac{1}{2}ln|\dfrac{1+cosx}{1-cosx}| \\ =-\dfrac{1}{2}ln\dfrac{(sinx)^2}{(1-cosx)^2} \\ =ln|\dfrac{1-cosx}{sinx}| \\ =ln|cscx + cotx| + C$

4

$\int (secx)^2\,dx=tanx + C,\quad tanx’=(secx)^2$
$\int (cscx)^2\,dx=-cotx +C,\quad cotx’=-(cscx)^2$

5

$\int secx \; tanx\, dx= \int \dfrac{sinx}{(cosx)^2}\,dx=\int-\dfrac{1}{(cosx)^2}\,d(cosx)=secx+C, \quad (secx)’=secx\;tanx$
$\int cscx \; cotx \,dx = \int\dfrac{cosx}{(sinx)^2}\,dx=\int\dfrac{1}{(sinx)^2}\,d(sinx)=-cscx + C, \quad (cscx)’ = cscx – cotx$

6

$\int\dfrac{1}{a^2 + x^2}\,dx=\dfrac{1}{a}arctan\dfrac{x}{a} + C$
$\int\dfrac{1}{\sqrt{a^2 – x^2}}\,dx=arcsin\dfrac{x}{a} + C$

7

三角函数换元

$\int\dfrac{1}{\sqrt{x^2 – a^2}}\,dx \\ 令\; x=asect, \quad dx=a\;sect \; tant \; dt \\ =\int \dfrac{1}{atant} \; asect \; tant\, dt \\ =\int sect\, dt \\ =ln|sect + tant| + C \\ 回代 \; x sect=\dfrac{x}{a} ,\quad tant=\dfrac{\sqrt{x^2 – a^2}}{a} \\ =ln | x + \sqrt{x^2 – a^2}| – lna + C_1 \\ =ln | x + \sqrt{x^2 – a^2}| + C$

$\int\dfrac{1}{\sqrt{x^2 + a^2}}\,dx \\ 令\; x=atant, \quad dx=a\;(sect)^2 \; dt \\ =\int \dfrac{1}{asect} \; a(sect)^2, dt \\ =\int sect\, dt \\ =ln|sect + tant| + C \\ 回代 \; sect = \dfrac{\sqrt{x^2 + a^2}}{a} \; tant=\dfrac{x}{a} \\ =ln|x + \sqrt{x^2 + a^2}| + C$

$\int\sqrt{a^2 – x^2}\, dx \\ 令\; x=asint \quad dx=acostdt \\ =\int a^2 (cost)^2\, dt \\ =a^2 \int \dfrac{cos2t+1}{2}\, dt \\ =a^2(\dfrac{sin2t}{4}+\dfrac{t}{2})+ C \\ 回代 \; sin2t=2sintcost=\dfrac{2x\sqrt{a^2-x^2}}{a^2} \quad t=arcsin\dfrac{x}{a} \\ =\dfrac{a^2}{2}arcsin\dfrac{x}{a} + \dfrac{x}{2}\sqrt{a^2 – x^2} + C$

8

平方差

$\int \dfrac{1}{x^2-a^2}\, dx \\ =\int \dfrac{1}{(x+a)(x-a)}\, dx \\ =\dfrac{1}{2a}[\int \dfrac{1}{x+a}\,dx – \int \dfrac{1}{x-a}\, dx]\\ =\dfrac{1}{2a}ln\dfrac{x+a}{x-a} + C$

9

三角函数变换

$\int (cosx)^2\, dx=\int \dfrac{cos2x+1}{2}\,dx=\dfrac{sin2x}{4}+\dfrac{x}{2}+ C$
$\int (sinx)^2\, dx=\int \dfrac{1-cos2x}{2}\, dx=\dfrac{x}{2} – \dfrac{sin2x}{4} + C$
$\int (tanx)^2\, dx=\int (secx)^2 – 1\, dx=tanx – x + C$
$\int (cotx)^2\, dx=\int (cscx)^2 – 1\, dx=-cotx – x + C$