⇢ 
Марк
АЗ
Андрей Задорожный
Здесь sin^(-1)=arcsin(x)
Take the integral:
integral (sin^(-1)(3 x))/sqrt(1-3 x) dx
For the integrand (sin^(-1)(3 x))/sqrt(1-3 x), substitute u = sqrt(1-3 x) and du = -3/(2 sqrt(1-3 x)) dx:
= -2/3 integral sin^(-1)(1-u^2) du
For the integrand sin^(-1)(1-u^2), integrate by parts, integral f dg = f g- integral g df, where
f = sin^(-1)(1-u^2), dg = du,
df = -(2 u)/sqrt(-(u^2 (u^2-2))) du, g = u:
= 2/3 integral -(2 u^2)/sqrt(-(u^2 (u^2-2))) du-2/3 u sin^(-1)(1-u^2)
Factor out constants:
= -2/3 u sin^(-1)(1-u^2)-4/3 integral u^2/sqrt(-(u^2 (u^2-2))) du
Expanding the integrand u^2/sqrt(-(u^2 (u^2-2))) gives u^2/sqrt(2 u^2-u^4):
= -2/3 u sin^(-1)(1-u^2)-4/3 integral u^2/sqrt(2 u^2-u^4) du
For the integrand u^2/sqrt(2 u^2-u^4), substitute s = u^2 and ds = 2 u du:
= -2/3 integral 1/sqrt(2-s) ds-2/3 u sin^(-1)(1-u^2)
For the integrand 1/sqrt(2-s), substitute p = 2-s and dp = - ds:
= 2/3 integral 1/sqrt(p) dp-2/3 u sin^(-1)(1-u^2)
The integral of 1/sqrt(p) is 2 sqrt(p):
= (4 sqrt(p))/3-2/3 u sin^(-1)(1-u^2)+constant
Substitute back for p = 2-s:
= (4 sqrt(2-s))/3-2/3 u sin^(-1)(1-u^2)+constant
Substitute back for s = u^2:
= (4 sqrt(2-u^2))/3-2/3 u sin^(-1)(1-u^2)+constant
Substitute back for u = sqrt(1-3 x):
= 4/3 sqrt(3 x+1)-2/3 sqrt(1-3 x) sin^(-1)(3 x)+constant
Factor the answer a different way:
= 1/3 (4 sqrt(3 x+1)-2 sqrt(1-3 x) sin^(-1)(3 x))+constant
Which is equivalent for restricted x values to:
Answer: |
| = (2 sqrt(1-3 x) ((1-3 x) sin^(-1)(3 x)-2 sqrt(1-9 x^2)))/(9 x-3)+constant
Похожие вопросы