math_三角升幂/降幂/微积分公式填空

三角函数

倍角

  1. s i n 2 x = sin2x= sin2x=
  2. c o s 2 x = cos2x= cos2x=
    1. 2f-1
    2. 1-2g
  3. t a n 2 x = tan2x= tan2x=

降幂(升角)公式

  1. s i n 2 x = sin^2x= sin2x=

  2. c o s 2 x = cos^2x= cos2x=

  3. 2 s i n 2 x 2 = 2sin^2\frac{x}{2}= 2sin22x=

  4. 2 c o s 2 x 2 = 2cos^2{\frac{x}{2}}= 2cos22x=

  5. 1 − c o s x = 1-cosx= 1cosx=

  6. s i n x c o s x = sinxcosx= sinxcosx=

  7. s i n x 2 c o s x 2 = sin\frac{x}{2}cos\frac{x}{2}= sin2xcos2x=

积分

  1. ∫ k d x = \int kdx= kdx=
  2. f 1 x d x = f \frac{1}{x}dx= fx1dx=
  3. ∫ s i n x d x = \int sinxdx= sinxdx=
  4. ∫ c o s x d x = \int cosxdx= cosxdx=
  5. ∫ x a d x = \int x^a dx= xadx=
  6. ∫ a x = \int a^x= ax=
  7. ∫ e x d x = \int e^xdx= exdx=
  8. ∫ 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=
  9. ∫ 1 s i n 2 = ∫ c s c 2 x d x = \int \frac{1}{sin^2}=\int csc^2xdx= sin21=csc2xdx=
  10. ∫ s e c x t a n x d x = \int secxtanxdx= secxtanxdx=
  11. ∫ c s c x c o t x d x = \int cscxcotxdx= cscxcotxdx=
  12. ∫ 1 1 − x 2 d x = \int \frac{1}{\sqrt{1-x^2}}dx= 1x2 1dx=
  13. ∫ 1 1 + x 2 d x = \int \frac{1}{1+x^2}dx= 1+x21dx=
  14. ∫ t a n x d x = \int tanxdx= tanxdx=
  15. ∫ c o t x d x = \int cotxdx= cotxdx=
  16. ∫ c s c x d x = \int cscxdx= cscxdx=
  17. ∫ s e c x d x = \int secxdx= secxdx=
  18. ∫ 1 a 2 + x 2 d x = \int \frac{1}{a^2+x^2}dx= a2+x21dx=
  19. ∫ 1 x 2 − a 2 d x = \int \frac{1}{x^2-a^2}dx= x2a21dx=
  20. ∫ 1 a 2 − x 2 d x = \int \frac{1}{\sqrt{a^2-x^2}}dx= a2x2 1dx=
  21. ∫ 1 x 2 + a 2 d x = \int \frac{1}{\sqrt{x^2+a^2}}dx= x2+a2 1dx=
  22. ∫ 1 x 2 − a 2 d x = \int \frac{1}{x^2-a^2}dx= x2a21dx=

参考答案

math_高数公式每日一过_part2(private)_xuchaoxin1375的博客-CSDN博客

积分

  1. ∫ k d x = k x + C \int kdx=kx+C kdx=kx+C

  2. ∫ d x = ∫ 1 d x = x + C \int dx=\int 1dx=x+C dx=1dx=x+C

  3. ∫ 0 d x = C \int 0dx=C 0dx=C

  4. f 1 x d x = l n ∣ x ∣ + C f \frac{1}{x}dx=ln|x|+C fx1dx=lnx+C

  5. ∫ s i n x d x = − c o s x + C \int sinxdx=-cosx+C sinxdx=cosx+C

  6. ∫ c o s x d x = s i n x + C \int cosxdx=sinx+C cosxdx=sinx+C

  7. ∫ 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

  8. ∫ 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)

  9. ∫ e x d x = e x + C \int e^xdx=e^x+C exdx=ex+C

  10. ∫ 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

  11. ∫ 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

  12. ∫ s e c x t a n x d x = s e c x + C \int secxtanxdx=secx+C secxtanxdx=secx+C

  13. ∫ c s c x c o t x d x = − c s c x + C \int cscxcotxdx=-cscx+C cscxcotxdx=cscx+C

  14. ∫ 1 1 − x 2 d x = a r c s i n x + C \int \frac{1}{\sqrt{1-x^2}}dx=arcsinx+C 1x2 1dx=arcsinx+C

  15. ∫ 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

  16. ∫ 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)=lncosx+C

  17. ∫ c o t x d x = ln ⁡ ∣ s i n x ∣ + C \int cotxdx=\ln |sinx|+C cotxdx=lnsinx+C

  18. ∫ c s c x d x = ln ⁡ ∣ c s c x − c o t x ∣ + C \int cscxdx=\ln |cscx-cotx|+C cscxdx=lncscxcotx+C

    1. 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
  19. ∫ s e c x d x = ln ⁡ ∣ s e c x + t a n x ∣ + C \int secxdx=\ln|secx+tanx|+C secxdx=lnsecx+tanx+C

  20. ∫ 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

  21. ∫ 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 x2a21dx=2a1lnx+axa+C

    1. 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x2a21列项后分项积分得到
  22. ∫ 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 a2x2 1dx=arcsin(ax)+C

  23. ∫ 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±a2 1dx=lnx+x2±a2 +C

    1. p = x 2 ± a 2 p=\sqrt{x^2\pm a^2} p=x2±a2
    2. ∫ 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 x2a2 1dx=p1dx=lnx+p+C
    3. 可由三角换元推导
  24. ∫ 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 a2x2 dx=2a2arcsinax+21a2x2 +C=21(a2arcsinax+a2x2 )+C

    1. p = a 2 − x 2 p=\sqrt{a^2-x^2} p=a2x2

    2. ∫ 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=21a2p1dx+21p+C=21(a2p1dx+p)+C

    3. 分部积分推导法

      1. 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=x2a2 dx=xx2a2 xdx2a2 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=xx2a2 =xdx2a2 =a2x2a2 1dx=a2lnx+x2a2

        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=xdx2a2 =x2a2 x2dx分子+0=a2+a2 x2a2 x2a2+a2dx=x2a2 dx+a2x2a2 1dx=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=AB=ASQ2S=AQS=21(AQ)S=21(xx2a2 a2lnx+x2a2 )

  25. ∫ 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+x2 dx=21(xa2+x2 +a2lna2+x2 +x)+C

    1. 利用三角换元配合分部积分法可以推导
    2. ∫ 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+lnsecx+tanx)+C
    3. 对于 S = ∫ a 2 ± x 2 d x 对于S=\int \sqrt{a^2\pm x^2}dx 对于S=a2±x2 dx
      1. p = a 2 ± x 2 p=\sqrt{a^2\pm x^2} p=a2±x2
      2. A = x p A=xp A=xp
      3. Q = a 2 ln ⁡ ∣ x + p ∣ Q=a^2\ln|x+p| Q=a2lnx+p
      4. S = 1 2 ( A ± Q ) + C S=\frac{1}{2}(A\pm Q)+C S=21(A±Q)+C
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