求反三角函数的原函数

Question

如题所述

Answer 1

用分部积分法得

I = ∫ arcsinx dx = x arcsinx - ∫ [x/√(1-x^2)] dx

= x arcsinx + (1/2) ∫ [1/√(1-x^2)] d(1-x^2) = x arcsinx + √(1-x^2) +C

I = ∫ arccosx dx = x arccosx + ∫ [x/√(1-x^2)] dx

= x arccosx - (1/2) ∫ [1/√(1-x^2)] d(1-x^2) = x arccosx - √(1-x^2) +C

I = ∫ arctanx dx = x arctanx - ∫ [x/(1+x^2)] dx

= x arctanx - (1/2) ∫ [1/(1+x^2)] d(1+x^2) = x arctanx - (1/2)ln(1+x^2) +C

扩展资料

不定积分的公式

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1

3、∫ 1/x dx = ln|x| + C

4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + C

6、∫ cosx dx = sinx + C

7、∫ sinx dx = - cosx + C

8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C

9、∫ tanx dx = - ln|cosx| + C = ln|secx| + C

10、∫ secx dx =ln|cot(x/2)| + C = (1/2)ln|(1 + sinx)/(1 - sinx)| + C = - ln|secx - tanx| + C = ln|secx + tanx| + C

Answer 2

那题说出来才能给你求啊，题说不出来怎么给你求。

Answer 3

用分部积分法得
I = ∫ arcsinx dx = x arcsinx - ∫ [x/√(1-x^2)] dx
= x arcsinx + (1/2) ∫ [1/√(1-x^2)] d(1-x^2) = x arcsinx + √(1-x^2) +C
I = ∫ arccosx dx = x arccosx + ∫ [x/√(1-x^2)] dx
= x arccosx - (1/2) ∫ [1/√(1-x^2)] d(1-x^2) = x arccosx - √(1-x^2) +C
I = ∫ arctanx dx = x arctanx - ∫ [x/(1+x^2)] dx
= x arctanx - (1/2) ∫ [1/(1+x^2)] d(1+x^2) = x arctanx - (1/2)ln(1+x^2) +C本回答被网友采纳