一.方型法

①:

a 1 c o s t b 1 s i n t a 1 s i n t + b 1 c o s t d t = l n a 1 s i n t + b 1 c o s t + C \int\frac{a_1cost-b_1sint}{a_1sint+b_1cost}dt=ln|a_1sint+b_1cost|+C a1sint+b1costa1costb1sintdt=lna1sint+b1cost+C
于是:
a 2 s i n t + b 2 c o s t a 1 s i n t + b 1 c o s t d t = A a 1 s i n t + b 1 c o s t a 1 s i n t + b 1 c o s t d t + B a 1 c o s t b 1 s i n t a 1 s i n t + b 1 c o s t d t \int\frac{a_2sint+b_2cost}{a_1sint+b_1cost}dt=A\int\frac{a_1sint+b_1cost}{a_1sint+b_1cost}dt+B\int\frac{a_1cost-b_1sint}{a_1sint+b_1cost}dt a1sint+b1costa2sint+b2costdt=Aa1sint+b1costa1sint+b1costdt+Ba1sint+b1costa1costb1sintdt
待定系数法求出 A , B A,B A,B
= A t + l n a 1 s i n t + b 1 c o s t + C 原式=At+ln|a_1sint+b_1cost|+C =At+lna1sint+b1cost+C

②:

1 1 + c o s t d t = 1 1 + c o s 2 t 2 s i n 2 t 2 d t = 1 2 c o s 2 t 2 d t = s e c 2 t 2 d t 2 = t a n t 2 \int\frac{1}{1+cost}dt=\int\frac{1}{1+cos^2{\frac{t}{2}}-sin^2{\frac{t}{2}}}dt=\int\frac{1}{2cos^2{\frac{t}{2}}}dt=\int sec^2\frac{t}{2}d\frac{t}{2}=tan\frac{t}{2} 1+cost1dt=1+cos22tsin22t1dt=2cos22t1dt=sec22td2t=tan2t

二.记忆型

①:

( s e c <mtext>   </mtext> t ) = s e c <mtext>   </mtext> t t a n <mtext>   </mtext> t (sec\ t)'=sec\ t\cdot tan\ t (sec t)=sec ttan t
( s e c t + t a n t ) = ( s e c 2 t + s e c t t a n t ) (sect+tant)'=(sec^2t+sect\cdot tant) (sect+tant)=(sec2t+secttant)
s e c t d t = l n s e c t + t a n t + C \int sectdt=ln|sect+tant|+C sectdt=lnsect+tant+C
推导:
s e c t <mtext>   </mtext> d t = s e c t ( s e c t + t a n t ) ( s e c t + t a n t ) d t = ( s e c 2 t + s e c t t a n t ) ( s e c t + t a n t ) d t = 1 s e c t + t a n t d ( s e c t + t a n t ) = l n s e c t + t a n t + C \int sect\ dt=\int\frac{sect(sect+tant)}{(sect+tant)}dt=\int\frac{(sec^2t+sect \cdot tant)}{(sect+tant)}dt=\int\frac{1}{sect+tant}d(sect+tant)=ln|sect+tant|+C sect dt=(sect+tant)sect(sect+tant)dt=(sect+tant)(sec2t+secttant)dt=sect+tant1d(sect+tant)=lnsect+tant+C

②:

( c s c t ) = c s c t c o t t (csct)'=-csct\cdot cot t (csct)=csctcott
( c s c + c o t ) = ( c s c 2 + c s c c o t ) (csc+cot)'=-(csc^2+csc\cdot cot) (csc+cot)=(csc2+csccot)
c s c t <mtext>   </mtext> d t = c s c t ( c s c t + c o t t ) ( c s c t + c o t t ) d t = ( c s c 2 t + c s c t c o t t ) ( c s c t + c o t t ) d t = 1 c s c t + c o t t d ( c s c t + c o t t ) = l n c s c t + c o t t \int csct\ dt=\int\frac{csct(csct+cott)}{(csct+cott)}dt=\int\frac{(csc^2t+csct \cdot cott)}{(csct+cott)}dt=\int\frac{-1}{csct+cott}d(csct+cott)=-ln|csct+cott| csct dt=(csct+cott)csct(csct+cott)dt=(csct+cott)(csc2t+csctcott)dt=csct+cott1d(csct+cott)=lncsct+cott

万能公式的那种

u = t a n x 2 u=tan\frac{x}{2} u=tan2x
x = 2 a r c t a n <mtext>   </mtext> u x=2arctan\ u x=2arctan u
d x = 2 1 + u 2 dx=\frac{2}{1+u^2} dx=1+u22

s i n x = 2 s i n x 2 c o s x 2 1 = = 2 s i n x 2 c o s x 2 s i n 2 x 2 + c o s 2 x 2 = 2 u 1 + u 2 sinx=\frac{2sin\frac{x}{2}cos\frac{x}{2}}{1}==\frac{2sin\frac{x}{2}cos\frac{x}{2}}{sin^2\frac{x}{2}+cos^2\frac{x}{2}}=\frac{2u}{1+u^2} sinx=12sin2xcos2x==sin22x+cos22x2sin2xcos2x=1+u22u

1 s i n x d x = l n ( t a n x 2 ) + C \int\frac{1}{sinx}dx=ln(tan\frac{x}{2})+C这个用万能公式比较好计算 sinx1dx=ln(tan2x)+C

③:
0 π x f ( s i n x ) d x = π 2 0 π f ( s i n x ) d x \int_0^\pi xf(sinx)dx=\frac{\pi}{2}\int_0^\pi f(sinx)dx 0πxf(sinx)dx=2π0πf(sinx)dx
证明:
I = 0 π x f ( s i n x ) d x = 0 π ( π x ) f ( s i n ( π x ) ) d x = 0 π ( π x ) f ( s i n x ) d x = π 0 π f ( s i n x ) I I=\int_0^\pi xf(sinx)dx=\int_0^\pi (\pi-x)f(sin(\pi-x))dx=\int_0^\pi (\pi-x)f(sinx)dx=\pi\int_0^{\pi}f(sinx)-I I=0πxf(sinx)dx=0π(πx)f(sin(πx))dx=0π(πx)f(sinx)dx=π0πf(sinx)I
I = π 2 0 π f ( s i n x ) \therefore I=\frac{\pi}{2}\int_0^{\pi}f(sinx) I=2π0πf(sinx)

三.完全记忆型

①:
1 a + x 2 d x = l n ( x + a + x 2 ) + C \int\frac{1}{\sqrt{a+x^2}}dx=ln(x+\sqrt{a+x^2})+C a+x2 1dx=ln(x+a+x2 )+C
②:
1 ( 1 + x 2 ) 3 2 d x = x 1 + x 2 + C \int\frac{1}{(1+x^2)^{\frac{3}{2}}}dx=\frac{x}{\sqrt{1+x^2}}+C (1+x2)231dx=1+x2 x+C
③:
1 x x 2 1 d x = a r c s i n 1 x + C \int\frac{1}{x\sqrt{x^2-1}}dx=-arcsin\frac{1}{x}+C xx21 1dx=arcsinx1+C
本来以为还要三角代换什么的很麻烦
1 x x 2 1 d x = 1 1 ( 1 x ) 2 d ( 1 x ) \int\frac{1}{x\sqrt{x^2-1}}dx=\int\frac{-1}{\sqrt{1-(\frac{1}{x})^2}}d(\frac{1}{x}) xx21 1dx=1(x1)2 1d(x1)
④:
s e c 3 x d x = 1 2 [ s e c x t a n x + s e c x d x ] \int sec^3xdx=\frac{1}{2}[secx\cdot tanx+\int secxdx] sec3xdx=21[secxtanx+secxdx]
I = s e c 3 x d x = s e c x d t a n x = s e c x t a n x t a n x d s e c x = s e c x t a n x t a n 2 x s e c x d x = s e c x t a n x ( s e c 2 x 1 ) s e c x d x = s e c x t a n x I + s e c x d x I = 1 2 [ s e c x t a n x + s e c x d x ] I=\int sec^3xdx=\int secxdtanx=secx\cdot tanx-\int tanxdsecx=secx\cdot tanx-\int tan^2xsecxdx=secx\cdot tanx-\int (sec^2x-1)secxdx=secx\cdot tanx-I+\int secxdx\Rightarrow I=\frac{1}{2}[secx\cdot tanx+\int secxdx] I=sec3xdx=secxdtanx=secxtanxtanxdsecx=secxtanxtan2xsecxdx=secxtanx(sec2x1)secxdx=secxtanxI+secxdxI=21[secxtanx+secxdx]

⑤:
a a c o s θ d θ = 2 2 a c o s θ 2 \int\sqrt{a-acos\theta}d\theta=-2\sqrt{2a}\cdot cos\frac{\theta}{2} aacosθ dθ=22a cos2θ
c o s θ 1 2 s i n 2 θ 2 cos\theta拆开成1-2sin^2\frac{\theta}{2} cosθ12sin22θ

a + a c o s θ d θ = 2 2 a s i n θ 2 \int\sqrt{a+acos\theta}d\theta=2\sqrt{2a}\cdot sin\frac{\theta}{2} a+acosθ dθ=22a sin2θ
变形成 c o s θ 2 cos\frac{\theta}{2} cos2θ所以积分后出来是正的 s i n sin sin,而第一种变成 s i n sin sin积分出来是负的 c o s cos cos
⑥:
1 s i n x c o s x d x = s e c 2 x t a n x d x = 1 t a n x d <mtext>   </mtext> t a n x = l n t a n x \int \frac{1}{sinxcosx}dx=\int\frac{sec^2x}{tanx}dx=\int\frac{1}{tanx}d\ tanx=ln|tanx| sinxcosx1dx=tanxsec2xdx=tanx1d tanx=lntanx
⑦:
1 1 + s i n x d x = 1 1 + 2 s i n x 2 c o s x 2 d x = 1 2 s i n 2 ( x 2 + π 4 ) d x = c s c 2 ( x 2 + π 4 ) d ( x 2 + π 4 ) = c o t ( x 2 + π 4 ) + C \int\frac{1}{1+sinx}dx=\int\frac{1}{1+2sin\frac{x}{2}cos\frac{x}{2}}dx=\int\frac{1}{2sin^2(\frac{x}{2}+\frac{\pi}{4})}dx=\int csc^2(\frac{x}{2}+\frac{\pi}{4})d(\frac{x}{2}+\frac{\pi}{4})=-cot(\frac{x}{2}+\frac{\pi}{4})+C 1+sinx1dx=1+2sin2xcos2x1dx=2sin2(2x+4π)1dx=csc2(2x+4π)d(2x+4π)=cot(2x+4π)+C
t a n ( x 2 π 4 ) = c o t ( x 2 + π 4 ) t a n : t a n ( x π 2 ) = c o t x , t a n s i n c o s 为啥tan(\frac{x}{2}-\frac{\pi}{4})=-cot(\frac{x}{2}+\frac{\pi}{4})这两个是相等的喃?好像是有个tan的诱导公式:tan(x-\frac{\pi}{2})=-cotx,这个把tan弄成\frac{sin}{cos}就能推出来 tan(2x4π)=cot(2x+4π)tan:tan(x2π)=cotx,tancossin
这样积分也是对的:
1 1 + s i n x d x = 1 s i n x 1 s i n 2 x d x = 1 s i n x c o s 2 x d x = s e c 2 x d x s i n x c o s 2 x d x = t a n x 1 c o s x + C \int\frac{1}{1+sinx}dx=\int\frac{1-sinx}{1-sin^2x}dx=\int\frac{1-sinx}{cos^2x}dx=\int sec^2xdx-\int \frac{sinx}{cos^2x}dx=tanx-\frac{1}{cosx}+C 1+sinx1dx=1sin2x1sinxdx=cos2x1sinxdx=sec2xdxcos2xsinxdx=tanxcosx1+C

傅里叶级数常用

{ <mstyle displaystyle="false" scriptlevel="0"> e a x cos n x d x = e a x a 2 + n 2 ( a c o s <mtext>   </mtext> n x + n s i n <mtext>   </mtext> n x ) </mstyle> <mstyle displaystyle="false" scriptlevel="0"> </mstyle> <mstyle displaystyle="false" scriptlevel="0"> e a x sin n x d x = e a x a 2 + n 2 ( a s i n n <mtext>   </mtext> x n c o s <mtext>   </mtext> n x ) </mstyle> \left\{\begin{matrix} \int e ^ { a x } \operatorname { cos } nx d x = \frac { e ^ { a x } } { a ^ { 2 } + n ^ { 2 } } ( acos\ nx + nsin\ nx )\\ \\\int e ^ { a x } \operatorname { sin } nx d x = \frac { e ^ { a x } } { a ^ { 2 } + n ^ { 2 } } ( asinn\ x - ncos\ nx ) \end{matrix}\right. eaxcosnxdx=a2+n2eax(acos nx+nsin nx)eaxsinnxdx=a2+n2eax(asinn xncos nx)

四.需要背的经典定积分

经典1

I = 0 e x 2 d x = π 2 I=\int_0^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2} I=0ex2dx=2π
是这样来的:
I I = 0 e x 2 d x 0 e x 2 d x = 0 e x 2 d x 0 e y 2 d y = 0 0 e ( x 2 + y 2 ) d x d y = 0 2 π 0 e r 2 r d r d θ r I\cdot I=\int_0^{\infty}e^{-x^2}dx\cdot \int_0^{\infty}e^{-x^2}dx=\int_0^{\infty}e^{-x^2}dx\cdot \int_0^{\infty}e^{-y^2}dy=\int_0^{\infty}\int_0^{\infty}e^{-(x^2+y^2)}dxdy=\int_0^{2\pi}\int_0^{\infty}e^{-r^2}rdrd\theta这里极坐标产生了个r就阔以拿进去积分了 II=0ex2dx0ex2dx=0ex2dx0ey2dy=00e(x2+y2)dxdy=02π0er2rdrdθr

华里士公式(点火公式)

I n = 0 π 2 s i n n x d x = 0 π 2 c o s n x d x = n 1 n I n 2 I_n=\int_0^{\frac{\pi}{2}}sin^nxdx=\int_0^{\frac{\pi}{2}}cos^nxdx=\frac{n-1}{n}I_{n-2} In=02πsinnxdx=02πcosnxdx=nn1In2
其中
I 0 = 0 π 2 s i n 0 x d x = 0 π 2 d x = π 2 I_0=\int_0^{\frac{\pi}{2}}sin^0xdx=\int_0^{\frac{\pi}{2}}dx=\frac{\pi}{2} I0=02πsin0xdx=02πdx=2π

I 1 = 0 π 2 s i n x d x = 1 I_1=\int_0^{\frac{\pi}{2}}sin^xdx=1 I1=02πsinxdx=1

常用等式

①:
a r c t a n A + a r c t a n 1 A = π 2 = a r c t a n A + a r c c o t A arctanA+arctan\frac{1}{A}=\frac{\pi}{2}=arctanA+arccotA arctanA+arctanA1=2π=arctanA+arccotA
: a r c t a n A = a , a r c t a n 1 A = b t a n a = A , t a n b = 1 A t a n a t a n b = 1 s i n a s i n b = c o s a c o s b c o s ( a + b ) = 0 a + b = π 2 设:arctanA=a,arctan\frac{1}{A}=b\Rightarrow tana=A,tanb=\frac{1}{A}\Rightarrow tana\cdot tanb=1\Rightarrow sina\cdot sinb=cosa\cdot cosb\Rightarrow cos(a+b)=0\Rightarrow a+b=\frac{\pi}{2} :arctanA=a,arctanA1=btana=A,tanb=A1tanatanb=1sinasinb=cosacosbcos(a+b)=0a+b=2π

做题遇到的比较怪的积分

①:
1 θ 2 + 1 θ 4 d θ = 1 θ 2 1 + θ 2 d θ , θ = t a n t , = c o t 2 t s e c 3 t d t = 1 s i n 2 t c o s t d t = s i n 2 t + c o s 2 t s i n 2 t c o s t d t = s e c t + c s c t c o t t <mtext>   </mtext> d t = l n s e c t + t a n t c s c t + C \int \sqrt{\frac{1}{\theta^2}+\frac{1}{\theta^4}}d\theta=\int \frac{1}{\theta^2}\sqrt{1+\theta^2}d\theta,令\theta=tant,=\int cot^2tsec^3tdt=\int\frac{1}{sin^2tcost}dt=\int\frac{sin^2t+cos^2t}{sin^2tcost}dt=\int sect+csct\cdot cott\ dt=ln|sect+tant|-csct+C θ21+θ41 dθ=θ211+θ2 dθ,θ=tant,=cot2tsec3tdt=sin2tcost1dt=sin2tcostsin2t+cos2tdt=sect+csctcott dt=lnsect+tantcsct+C

比较好的积分题

2017版张宇1000题的

3.88

x ( y x ) 2 = y , 1 y x x(y-x)^2=y,求\int\frac{1}{y-x} x(yx)2=y,yx1
要令 y x = t y-x=t yx=t

3.120

1 x 6 ( 1 + x 2 ) d x 求\int\frac{1}{x^6(1+x^2)}dx x6(1+x2)1dx
技巧: 1 = ( 1 + x 2 ) x 2 1=(1+x^2)-x^2 1=(1+x2)x2

3.121

0 π 2 d x 1 + ( t a n x ) 2 求\int_0^{\frac{\pi}{2}}\frac{dx}{1+(tanx)^{\sqrt{2}}} 02π1+(tanx)2 dx
看看 f ( π 2 x ) + f ( x ) f(\frac{\pi}{2}-x)+f(x) f(2πx)+f(x)会有什么结果?

3.122

0 π 1 1 + a c o s <mtext>   </mtext> x d x \int_0^{\pi}\frac{1}{1+acos\ x}dx 0π1+acos x1dx
分成 0 π 2 + π 2 π \int_0^{\frac{\pi}{2}}+\int_\frac{\pi}{2}^{\pi} 02π+2ππ两部分,并且把后面那个换元,使得上下限也是 0 π 2 \int_0^{\frac{\pi}{2}} 02π,然后看会发生什么

3.123

0 + s i n x x d x = π 2 0 + s i n 2 x x 2 d x 已知\int_0^{+\infty}\frac{sinx}{x}dx=\frac{\pi}{2},求\int_0^{+\infty}\frac{sin^2x}{x^2}dx 0+xsinxdx=2π0+x2sin2xdx
I ( a ) = 0 + s i n 2 a x x 2 d x I(a)=\int_0^{+\infty}\frac{sin^2ax}{x^2}dx I(a)=0+x2sin2axdx
会发现 d I ( a ) d a \frac{dI(a)}{da} dadI(a) 0 + s i n x x d x = π 2 \int_0^{+\infty}\frac{sinx}{x}dx=\frac{\pi}{2} 0+xsinxdx=2π长得有点像

2020版张宇1000题的

3.31

f ( x ) = e x + e x 2 , [ f ( x ) f ( x ) + f ( x ) f ( x ) ] d x f(x)=\frac{e^x+e^{-x}}{2},求\int[\frac{f'(x)}{f(x)}+\frac{f(x)}{f'(x)}]dx f(x)=2ex+ex,[f(x)f(x)+f(x)f(x)]dx
如果能发现 f ( x ) = f ( x ) f''(x)=f(x) f(x)=f(x)就好做了

3.104

I = 0 1 x b x a l n x d x , ( a , b > 0 ) 求I=\int_0^1\frac{x^b-x^a}{lnx}dx,其中(a,b>0) I=01lnxxbxadx,(a,b>0)
又是一道阔以拿去坑大林的积分题~嘿嘿嘿,看他做过没
竟然是化成二重积分来算的T_T,然后还要换次序才能算出来
I = 0 1 a b x y d y d x = a b 0 1 x y d x d y = a b 1 1 + y d y = l n 1 + b 1 + a I=\int_0^1\int_a^bx^ydydx=\int_a^b\int_0^1x^ydxdy=\int_a^b\frac{1}{1+y}dy=ln\frac{1+b}{1+a} I=01abxydydx=ab01xydxdy=ab1+y1dy=ln1+a1+b

3.105

I = 0 + 1 ( 1 + x 2 ) ( 1 + x α ) d x , α 0 求I=\int_0^{+\infty}\frac{1}{(1+x^2)(1+x^{\alpha})}dx,其中\alpha不等于0 I=0+(1+x2)(1+xα)1dx,α0
妙啊~
, t = 1 x , I = 0 + t α ( 1 + t 2 ) ( 1 + t α ) d t 换元,令t=\frac{1}{x},I=\int_0^{+\infty}\frac{t^{\alpha}}{(1+t^2)(1+t^{\alpha})}dt ,t=x1,I=0+(1+t2)(1+tα)tαdt

I + I = 0 + 1 + x α ( 1 + x 2 ) ( 1 + x α ) d x = 0 + 1 ( 1 + x 2 ) d x = a r c t a n ( ) a r c t a n ( 0 ) = π 2 \therefore I+I=\int_0^{+\infty}\frac{1+x^{\alpha}}{(1+x^2)(1+x^{\alpha})}dx=\int_0^{+\infty}\frac{1}{(1+x^2)}dx=arctan(\infty)-arctan(0)=\frac{\pi}{2} I+I=0+(1+x2)(1+xα)1+xαdx=0+(1+x2)1dx=arctan()arctan(0)=2π

I = π 4 \therefore I=\frac{\pi}{4} I=4π

其他地方看到的

1

0 1 x b x a l n x d x \int_0^1\frac{x^b-x^a}{ln x}dx 01lnxxbxadx
这道题是在b站一位叫:MATH-IDEA的up主看到的,感谢分享(✪ω✪)

重积分法

主要是对指数的积分不熟, a x d x = a x l n a \int a^xdx=\frac{a^x}{ln a} axdx=lnaax,好像就没怎么对指数积过分样,都是求导
0 1 x b x a l n x d x = 0 1 a b x y d y d x \int_0^1\frac{x^b-x^a}{ln x}dx=\int_0^1\int_a^b x^ydydx然后交换次序 01lnxxbxadx=01abxydydx

求导法

F ( t ) = 0 1 x t x a l n x d x F(t)=\int_0^1\frac{x^t-x^a}{ln x}dx F(t)=01lnxxtxadx
F ( t ) = <munder> lim t 0 </munder> F ( t + t ) F ( t ) t = <munder> lim t 0 </munder> 0 1 [ x t + t l n x x t l n x ] d x t = <munder> lim t 0 </munder> 0 1 [ x t + t x t t ] 1 l n x d x l n x = 0 1 x t d x = 1 t + 1 x t + 1 0 1 = 1 t + 1 F'(t)=\lim_{\triangle t\to0}\frac{F(t+\triangle t)-F(t)}{\triangle t}代入化简=\lim_{\triangle t\to0}\frac{\int_0^1[\frac{x^{t+\triangle t}}{lnx}-\frac{x^t}{lnx}]dx}{\triangle t}=\lim_{\triangle t\to0}\int_0^1[\frac{x^{t+\triangle t}-x^{t}}{\triangle t}]\cdot\frac{1}{lnx}dx里面打括号的就是指数函数的导数,然后与lnx约掉=\int_0^1x^tdx=\frac{1}{t+1}x^{t+1}|_0^1=\frac{1}{t+1} F(t)=t0limtF(t+t)F(t)=t0limt01[lnxxt+tlnxxt]dx=t0lim01[txt+txt]lnx1dxlnx=01xtdx=t+11xt+101=t+11

所以就得到了: F ( t ) = 1 t + 1 F'(t)=\frac{1}{t+1} F(t)=t+11
再积分回来: F ( t ) = l n ( t + 1 ) + C F(t)=ln(t+1)+C F(t)=ln(t+1)+C

F ( a ) = 0 C = l n ( a + 1 ) F(a)=0\Rightarrow C=-ln(a+1) F(a)=0C=ln(a+1)
然后所求的就是 F ( b ) F(b) F(b)

换元法

需要知道一个结论:傅汝兰尼积分
0 f ( b x ) f ( a x ) x d x = [ f ( 0 ) f ( ) ] l n b a \int_{-\infty}^0\frac{f(bx)-f(ax)}{x}dx=[f(0)-f(-\infty)]\cdot ln\frac{b}{a} 0xf(bx)f(ax)dx=[f(0)f()]lnab
t = l n x t=lnx t=lnx
换元之后就是:
0 e ( b + 1 ) t e ( a + 1 ) t t d t \int_{-\infty}^0\frac{e^{(b+1)t}-e^{(a+1)t}}{t}dt 0te(b+1)te(a+1)tdt