大家好，欢迎来到IT知识分享网。
三角函数
倍角
- s i n 2 x = sin2x= sin2x=
- c o s 2 x = cos2x= cos2x=
2f-11-2g
- t a n 2 x = tan2x= tan2x=
降幂(升角)公式
- s i n 2 x = sin^2x= sin2x=
- c o s 2 x = cos^2x= cos2x=
- 2 s i n 2 x 2 = 2sin^2\frac{x}{2}= 2sin22x=
- 2 c o s 2 x 2 = 2cos^2{\frac{x}{2}}= 2cos22x=
- 1 − c o s x = 1-cosx= 1−cosx=
- s i n x c o s x = sinxcosx= sinxcosx=
- s i n x 2 c o s x 2 = sin\frac{x}{2}cos\frac{x}{2}= sin2xcos2x=
积分
- ∫ k d x = \int kdx= ∫kdx=
- f 1 x d x = f \frac{1}{x}dx= fx1dx=
- ∫ s i n x d x = \int sinxdx= ∫sinxdx=
- ∫ c o s x d x = \int cosxdx= ∫cosxdx=
- ∫ x a d x = \int x^a dx= ∫xadx=
- ∫ a x = \int a^x= ∫ax=
- ∫ e x d x = \int e^xdx= ∫exdx=
- ∫ 1 c o s 2 x d x = ∫ s e c 2 x d x = \int \frac{1}{cos^2x}dx=\int sec^2xdx= ∫cos2x1dx=∫sec2xdx=
- ∫ 1 s i n 2 = ∫ c s c 2 x d x = \int \frac{1}{sin^2}=\int csc^2xdx= ∫sin21=∫csc2xdx=
- ∫ s e c x t a n x d x = \int secxtanxdx= ∫secxtanxdx=
- ∫ c s c x c o t x d x = \int cscxcotxdx= ∫cscxcotxdx=
- ∫ 1 1 − x 2 d x = \int \frac{1}{\sqrt{1-x^2}}dx= ∫1−x21dx=
- ∫ 1 1 + x 2 d x = \int \frac{1}{1+x^2}dx= ∫1+x21dx=
- ∫ t a n x d x = \int tanxdx= ∫tanxdx=
- ∫ c o t x d x = \int cotxdx= ∫cotxdx=
- ∫ c s c x d x = \int cscxdx= ∫cscxdx=
- ∫ s e c x d x = \int secxdx= ∫secxdx=
- ∫ 1 a 2 + x 2 d x = \int \frac{1}{a^2+x^2}dx= ∫a2+x21dx=
- ∫ 1 x 2 − a 2 d x = \int \frac{1}{x^2-a^2}dx= ∫x2−a21dx=
- ∫ 1 a 2 − x 2 d x = \int \frac{1}{\sqrt{a^2-x^2}}dx= ∫a2−x21dx=
- ∫ 1 x 2 + a 2 d x = \int \frac{1}{\sqrt{x^2+a^2}}dx= ∫x2+a21dx=
- ∫ 1 x 2 − a 2 d x = \int \frac{1}{x^2-a^2}dx= ∫x2−a21dx=
参考答案
math_高数公式每日一过_part2(private)_xuchaoxin1375的博客-CSDN博客
积分
- ∫ k d x = k x + C \int kdx=kx+C ∫kdx=kx+C
- ∫ d x = ∫ 1 d x = x + C \int dx=\int 1dx=x+C ∫dx=∫1dx=x+C
- ∫ 0 d x = C \int 0dx=C ∫0dx=C
- f 1 x d x = l n ∣ x ∣ + C f \frac{1}{x}dx=ln|x|+C fx1dx=ln∣x∣+C
- ∫ s i n x d x = − c o s x + C \int sinxdx=-cosx+C ∫sinxdx=−cosx+C
- ∫ c o s x d x = s i n x + C \int cosxdx=sinx+C ∫cosxdx=sinx+C
- ∫ x a d x = 1 a + 1 x a + 1 \int x^a dx=\frac{1}{a+1}x^{a+1} ∫xadx=a+11xa+1
- ∫ a x = a x ln a + C ( a > 0 ; a ≠ 1 ) \int a^x=\frac{a^x}{\ln a}+C(a>0;a\ne1) ∫ax=lnaax+C(a>0;a=1)
- ∫ e x d x = e x + C \int e^xdx=e^x+C ∫exdx=ex+C
- ∫ 1 c o s 2 x d x = ∫ s e c 2 x d x = t a n x + C \int \frac{1}{cos^2x}dx=\int sec^2xdx=tanx+C ∫cos2x1dx=∫sec2xdx=tanx+C
- ∫ 1 s i n 2 = ∫ c s c 2 x d x = − c o t x + C \int \frac{1}{sin^2}=\int csc^2xdx=-cotx+C ∫sin21=∫csc2xdx=−cotx+C
- ∫ s e c x t a n x d x = s e c x + C \int secxtanxdx=secx+C ∫secxtanxdx=secx+C
- ∫ c s c x c o t x d x = − c s c x + C \int cscxcotxdx=-cscx+C ∫cscxcotxdx=−cscx+C
- ∫ 1 1 − x 2 d x = a r c s i n x + C \int \frac{1}{\sqrt{1-x^2}}dx=arcsinx+C ∫1−x21dx=arcsinx+C
- ∫ 1 1 + x 2 d x = a r c t a n x + C \int \frac{1}{1+x^2}dx=arctanx+C ∫1+x21dx=arctanx+C
- ∫ t a n x d x = ∫ s i n x c o s x d x = ∫ d ( − c o s x ) s i n x = − ln ∣ c o s x ∣ + C \int tanxdx=\int \frac{sinx}{cosx}dx=\int \frac{d(-cosx)}{sinx}=-\ln|cosx|+C ∫tanxdx=∫cosxsinxdx=∫sinxd(−cosx)=−ln∣cosx∣+C
- ∫ c o t x d x = ln ∣ s i n x ∣ + C \int cotxdx=\ln |sinx|+C ∫cotxdx=ln∣sinx∣+C
- ∫ c s c x d x = ln ∣ c s c x − c o t x ∣ + C \int cscxdx=\ln |cscx-cotx|+C ∫cscxdx=ln∣cscx−cotx∣+C
- 可由三角降角升幂和配凑乘以 1 = c o s x c o s x 可由三角降角升幂和配凑乘以1=\frac{cosx}{cosx} 可由三角降角升幂和配凑乘以1=cosxcosx
- ∫ s e c x d x = ln ∣ s e c x + t a n x ∣ + C \int secxdx=\ln|secx+tanx|+C ∫secxdx=ln∣secx+tanx∣+C
- ∫ 1 a 2 + x 2 d x = 1 a a r c t a n x a + C \int \frac{1}{a^2+x^2}dx=\frac{1}{a}arctan{\frac{x}{a}}+C ∫a2+x21dx=a1arctanax+C
- ∫ 1 x 2 − a 2 d x = 1 2 a ln ∣ x − a x + a ∣ + C \int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C ∫x2−a21dx=2a1ln∣x+ax−a∣+C
- 可由 1 x 2 − a 2 列项后分项积分得到 可由\frac{1}{x^2-a^2}列项后分项积分得到 可由x2−a21列项后分项积分得到
- ∫ 1 a 2 − x 2 d x = a r c s i n ( x a ) + C \int \frac{1}{\sqrt{a^2-x^2}}dx=arcsin(\frac{x}{a})+C ∫a2−x21dx=arcsin(ax)+C
- ∫ 1 x 2 ± a 2 d x = ln ∣ x + x 2 ± a 2 ∣ + C \int \frac{1}{\sqrt{x^2\pm a^2}}dx=\ln|x+\sqrt{x^2\pm a^2}|+C ∫x2±a21dx=ln∣x+x2±a2∣+C
- p = x 2 ± a 2 p=\sqrt{x^2\pm a^2} p=x2±a2
- ∫ 1 x 2 − a 2 d x = ∫ 1 p d x = ln ∣ x + p ∣ + C \int \frac{1}{\sqrt{x^2-a^2}}dx=\int \frac{1}{p}dx=\ln|x+p|+C ∫x2−a21dx=∫p1dx=ln∣x+p∣+C
- 可由三角换元推导
- ∫ a 2 − x 2 d x = a 2 2 a r c s i n x a + 1 2 a 2 − x 2 + C = 1 2 ( a 2 a r c s i n x a + a 2 − x 2 ) + C \int \sqrt{a^2-x^2}dx=\frac{a^2}{2}arcsin{\frac{x}{a}}+\frac{1}{2}\sqrt{a^2-x^2}+C=\frac{1}{2}(a^2arcsin{\frac{x}{a}}+\sqrt{a^2-x^2})+C ∫a2−x2dx=2a2arcsinax+21a2−x2+C=21(a2arcsinax+a2−x2)+C
- p = a 2 − x 2 p=\sqrt{a^2-x^2} p=a2−x2
- ∫ p d x = 1 2 a 2 ∫ 1 p d x + 1 2 p + C = 1 2 ( a 2 ∫ 1 p d x + p ) + C \int pdx=\frac{1}{2}a^2\int \frac{1}{p}dx+\frac{1}{2}p+C=\frac{1}{2}(a^2\int \frac{1}{p}dx+p)+C ∫pdx=21a2∫p1dx+21p+C=21(a2∫p1dx+p)+C
- 分部积分推导法
- S = ∫ x 2 − a 2 d x = x x 2 − a 2 − ∫ x d x 2 − a 2 为例方便说明推导和简洁性 , 提前给出如下标记 ( 表达式记号 ) A = x x 2 − a 2 B = ∫ x d x 2 − a 2 Q = a 2 ∫ 1 x 2 − a 2 d x = a 2 ln ∣ x + x 2 − a 2 ∣ \\ \begin{aligned} S&=\int \sqrt{x^2-a^2}dx \\ &=x\sqrt{x^2-a^2}-\int xd\sqrt{x^2-a^2} \end{aligned} \\为例方便说明推导和简洁性,提前给出如下标记(表达式记号) \\ \begin{aligned} \\A&=x\sqrt{x^2-a^2} \\B&=\int xd\sqrt{x^2-a^2} \\Q&=a^2\int\frac{1}{\sqrt{x^2-a^2}}dx=a^2\ln |x+\sqrt{x^2-a^2}| \end{aligned} S=∫x2−a2dx=xx2−a2−∫xdx2−a2为例方便说明推导和简洁性,提前给出如下标记(表达式记号)ABQ=xx2−a2=∫xdx2−a2=a2∫x2−a21dx=a2ln∣x+x2−a2∣
B = ∫ x d x 2 − a 2 = ∫ x 2 x 2 − a 2 d x = 分子 + 0 = − a 2 + a 2 ∫ x 2 − a 2 + a 2 x 2 − a 2 d x = ∫ x 2 − a 2 d x + a 2 ∫ 1 x 2 − a 2 d x = S + Q \begin{aligned} B &=\int xd\sqrt{x^2-a^2}\\ &=\int \frac{x^2}{\sqrt{x^2-a^2}}dx \\ &\xlongequal{分子+0=-a^2+a^2}\int \frac{x^2-a^2+a^2}{\sqrt{x^2-a^2}}dx\\ &=\int\sqrt{x^2-a^2}dx+a^2\int\frac{1}{\sqrt{x^2-a^2}}dx\\ \\ &=S+Q \end{aligned} B=∫xdx2−a2=∫x2−a2x2dx分子+0=−a2+a2∫x2−a2x2−a2+a2dx=∫x2−a2dx+a2∫x2−a21dx=S+Q
S = A − B = A − S − Q 2 S = A − Q → S = 1 2 ( A − Q ) S = 1 2 ( x x 2 − a 2 − a 2 ln ∣ x + x 2 − a 2 ∣ ) \\S=A-B=A-S-Q \\2S=A-Q \to S=\frac{1}{2}(A-Q) \\S=\frac{1}{2}(x\sqrt{x^2-a^2}-a^2\ln |x+\sqrt{x^2-a^2}|) S=A−B=A−S−Q2S=A−Q→S=21(A−Q)S=21(xx2−a2−a2ln∣x+x2−a2∣)
- S = ∫ x 2 − a 2 d x = x x 2 − a 2 − ∫ x d x 2 − a 2 为例方便说明推导和简洁性 , 提前给出如下标记 ( 表达式记号 ) A = x x 2 − a 2 B = ∫ x d x 2 − a 2 Q = a 2 ∫ 1 x 2 − a 2 d x = a 2 ln ∣ x + x 2 − a 2 ∣ \\ \begin{aligned} S&=\int \sqrt{x^2-a^2}dx \\ &=x\sqrt{x^2-a^2}-\int xd\sqrt{x^2-a^2} \end{aligned} \\为例方便说明推导和简洁性,提前给出如下标记(表达式记号) \\ \begin{aligned} \\A&=x\sqrt{x^2-a^2} \\B&=\int xd\sqrt{x^2-a^2} \\Q&=a^2\int\frac{1}{\sqrt{x^2-a^2}}dx=a^2\ln |x+\sqrt{x^2-a^2}| \end{aligned} S=∫x2−a2dx=xx2−a2−∫xdx2−a2为例方便说明推导和简洁性,提前给出如下标记(表达式记号)ABQ=xx2−a2=∫xdx2−a2=a2∫x2−a21dx=a2ln∣x+x2−a2∣
- ∫ a 2 + x 2 d x = 1 2 ( x a 2 + x 2 + a 2 ln ∣ a 2 + x 2 + x ∣ ) + C \int \sqrt{a^2+x^2}dx=\frac{1}{2}(x\sqrt{a^2+x^2}+a^2 \ln|\sqrt{a^2+x^2}+x|) +C ∫a2+x2dx=21(xa2+x2+a2ln∣a2+x2+x∣)+C
- 利用三角换元配合分部积分法可以推导
- ∫ s e c 3 t d t = 1 2 ( s e c x t a n x + ln ∣ s e c x + t a n x ∣ ) + C \int sec^3t\ dt=\frac{1}{2}(secxtanx+\ln |secx+tanx|)+C ∫sec3t dt=21(secxtanx+ln∣secx+tanx∣)+C
- 对于 S = ∫ a 2 ± x 2 d x 对于S=\int \sqrt{a^2\pm x^2}dx 对于S=∫a2±x2dx
- p = a 2 ± x 2 p=\sqrt{a^2\pm x^2} p=a2±x2
- A = x p A=xp A=xp
- Q = a 2 ln ∣ x + p ∣ Q=a^2\ln|x+p| Q=a2ln∣x+p∣
- S = 1 2 ( A ± Q ) + C S=\frac{1}{2}(A\pm Q)+C S=21(A±Q)+C
免责声明：本站所有文章内容,图片，视频等均是来源于用户投稿和互联网及文摘转载整编而成，不代表本站观点，不承担相关法律责任。其著作权各归其原作者或其出版社所有。如发现本站有涉嫌抄袭侵权/违法违规的内容,侵犯到您的权益，请在线联系站长,一经查实,本站将立刻删除。 本文来自网络,若有侵权，请联系删除，如若转载，请注明出处：https://haidsoft.com/148604.html
