不定积分基本公式与推导

1

∫ax dx=axlna+C,(ax)′=axlna\int\ a^x\,dx=\dfrac{a^x}{lna} + C,\quad ({a^x})’ = a^x lna∫axdx=lnaax​+C,(ax)′=axlna

2

∫tan⁡x dx=∫sinxcosx dx=−∫1cosx d(cosx)=−ln⁡∣cos⁡x∣+C\int\tan x\,dx=\int\dfrac{sinx}{cosx}\,dx=-\int\dfrac{1}{cosx}\,d(cosx)=-\ln{|\cos{x}|} + C∫tanxdx=∫cosxsinx​dx=−∫cosx1​d(cosx)=−ln∣cosx∣+C

∫cot⁡x dx=∫cosxsinx dx=−∫1sinx d(sinx)=ln⁡∣sin⁡x∣+C\int\cot x\,dx=\int\dfrac{cosx}{sinx}\,dx=-\int\dfrac{1}{sinx}\,d(sinx)=\ln{|\sin{x}|} + C∫cotxdx=∫sinxcosx​dx=−∫sinx1​d(sinx)=ln∣sinx∣+C

3

∫1cosx dx=∫sec⁡x dx=∫cosx(cosx)2 dx=∫cosx1−(sinx)2 dx=∫11−(sinx)2 d(sinx)=∫1(1+sinx)(1−sinx) d(sinx)=12[∫11+sinx d(sinx)−∫11−sinx d(sinx)]=12ln∣1+sinx1−sinx∣+C=12ln∣(1+sinx)2(cosx)2∣=ln∣1+sinxcosx∣=ln∣secx+tanx∣+C\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∫cosx1​dx=∫secxdx=∫(cosx)2cosx​dx=∫1−(sinx)2cosx​dx=∫1−(sinx)21​d(sinx)=∫(1+sinx)(1−sinx)1​d(sinx)=21​[∫1+sinx1​d(sinx)−∫1−sinx1​d(sinx)]=21​ln∣1−sinx1+sinx​∣+C=21​ln∣(cosx)2(1+sinx)2​∣=ln∣cosx1+sinx​∣=ln∣secx+tanx∣+C

∫1sinx dx=∫csc⁡x dx=∫sinx(sinx)2 dx=∫sinx1−(cosx)2 dx=−∫11−(cosx)2 d(cosx)=−∫1(1+cosx)(1−cosx) d(cosx)=−12[∫11+cosx d(cosx+1)−∫11−cosx d(1−cosx)]=−12ln∣1+cosx1−cosx∣=−12ln(sinx)2(1−cosx)2=ln∣1−cosxsinx∣=ln∣cscx+cotx∣+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∫sinx1​dx=∫cscxdx=∫(sinx)2sinx​dx=∫1−(cosx)2sinx​dx=−∫1−(cosx)21​d(cosx)=−∫(1+cosx)(1−cosx)1​d(cosx)=−21​[∫1+cosx1​d(cosx+1)−∫1−cosx1​d(1−cosx)]=−21​ln∣1−cosx1+cosx​∣=−21​ln(1−cosx)2(sinx)2​=ln∣sinx1−cosx​∣=ln∣cscx+cotx∣+C

4

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

5

∫secx  tanx dx=∫sinx(cosx)2 dx=∫−1(cosx)2 d(cosx)=secx+C,(secx)′=secx  tanx\int secx \; tanx\, dx= \int \dfrac{sinx}{(cosx)^2}\,dx=\int-\dfrac{1}{(cosx)^2}\,d(cosx)=secx+C, \quad (secx)’=secx\;tanx∫secxtanxdx=∫(cosx)2sinx​dx=∫−(cosx)21​d(cosx)=secx+C,(secx)′=secxtanx
∫cscx  cotx dx=∫cosx(sinx)2 dx=∫1(sinx)2 d(sinx)=−cscx+C,(cscx)′=cscx−cotx\int cscx \; cotx \,dx = \int\dfrac{cosx}{(sinx)^2}\,dx=\int\dfrac{1}{(sinx)^2}\,d(sinx)=-cscx + C, \quad (cscx)’ = cscx – cotx∫cscxcotxdx=∫(sinx)2cosx​dx=∫(sinx)21​d(sinx)=−cscx+C,(cscx)′=cscx−cotx

6

∫1a2+x2 dx=1aarctanxa+C\int\dfrac{1}{a^2 + x^2}\,dx=\dfrac{1}{a}arctan\dfrac{x}{a} + C∫a2+x21​dx=a1​arctanax​+C
∫1a2−x2 dx=arcsinxa+C\int\dfrac{1}{\sqrt{a^2 – x^2}}\,dx=arcsin\dfrac{x}{a} + C∫a2−x2​1​dx=arcsinax​+C

7

三角函数换元

∫1×2−a2 dx令  x=asect,dx=a  sect  tant  dt=∫1atant  asect  tant dt=∫sect dt=ln∣sect+tant∣+C回代  xsect=xa,tant=x2−a2a=ln∣x+x2−a2∣−lna+C1=ln∣x+x2−a2∣+C\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 ∫x2−a2​1​dx令x=asect,dx=asecttantdt=∫atant1​asecttantdt=∫sectdt=ln∣sect+tant∣+C回代xsect=ax​,tant=ax2−a2​​=ln∣x+x2−a2​∣−lna+C1​=ln∣x+x2−a2​∣+C

∫1×2+a2 dx令  x=atant,dx=a  (sect)2  dt=∫1asect  a(sect)2,dt=∫sect dt=ln∣sect+tant∣+C回代  sect=x2+a2a  tant=xa=ln∣x+x2+a2∣+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 ∫x2+a2​1​dx令x=atant,dx=a(sect)2dt=∫asect1​a(sect)2,dt=∫sectdt=ln∣sect+tant∣+C回代sect=ax2+a2​​tant=ax​=ln∣x+x2+a2​∣+C

∫a2−x2 dx令  x=asintdx=acostdt=∫a2(cost)2 dt=a2∫cos2t+12 dt=a2(sin2t4+t2)+C回代  sin2t=2sintcost=2xa2−x2a2t=arcsinxa=a22arcsinxa+x2a2−x2+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∫a2−x2​dx令x=asintdx=acostdt=∫a2(cost)2dt=a2∫2cos2t+1​dt=a2(4sin2t​+2t​)+C回代sin2t=2sintcost=a22xa2−x2​​t=arcsinax​=2a2​arcsinax​+2x​a2−x2​+C

8

平方差

∫1×2−a2 dx=∫1(x+a)(x−a) dx=12a[∫1x+a dx−∫1x−a dx]=12alnx+ax−a+C
\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∫x2−a21​dx=∫(x+a)(x−a)1​dx=2a1​[∫x+a1​dx−∫x−a1​dx]=2a1​lnx−ax+a​+C

9

三角函数改换

∫(cosx)2 dx=∫cos2x+12 dx=sin2x4+x2+C\int (cosx)^2\, dx=\int \dfrac{cos2x+1}{2}\,dx=\dfrac{sin2x}{4}+\dfrac{x}{2}+ C∫(cosx)2dx=∫2cos2x+1​dx=4sin2x​+2x​+C
∫(sinx)2 dx=∫1−cos2x2 dx=x2−sin2x4+C\int (sinx)^2\, dx=\int \dfrac{1-cos2x}{2}\, dx=\dfrac{x}{2} – \dfrac{sin2x}{4} + C∫(sinx)2dx=∫21−cos2x​dx=2x​−4sin2x​+C
∫(tanx)2 dx=∫(secx)2−1 dx=tanx−x+C\int (tanx)^2\, dx=\int (secx)^2 – 1\, dx=tanx – x + C∫(tanx)2dx=∫(secx)2−1dx=tanx−x+C
∫(cotx)2 dx=∫(cscx)2−1 dx=−cotx−x+C\int (cotx)^2\, dx=\int (cscx)^2 – 1\, dx=-cotx – x + C∫(cotx)2dx=∫(cscx)2−1dx=−cotx−x+C