Question

计算不定积分∫xarcsinxdx

Answer 1

∫xarcsinxdx
=1/2∫arcsinxdx²
=1/2x²*arcsinx-1/2∫x²darcsinx
=1/2x²*arcsinx-1/2∫x²/√(1-x²)dx
=1/2x²*arcsinx+1/2∫-x²/√(1-x²)dx
=1/2x²*arcsinx+1/2∫(1-x²-1)/√(1-x²)dx
=1/2x²*arcsinx+1/2∫[(1-x²)/√(1-x²)-1/√(1-x²)]dx
=1/2x²*arcsinx+1/2∫[√(1-x²)-1/√(1-x²)]dx
=1/2x²*arcsinx+1/2∫√(1-x²)dx-arcsinx
单独求∫√(1-x²)dx
令x=sina
√(1-x²)=cosa
sin2a=2sinacosa=2x√(1-x²)
dx=cosada
∫√(1-x²)dx
=∫cosa*cosada
=∫(1+cos2a)/2 da
=1/2∫da+1/4∫cos2ad2a
=a/2+sin2a/4
=arcsinx/2+2x√(1-x²)/4
=arcsinx/2+x√(1-x²)/2
所以原式=1/2x²*arcsinx+(arcsinx)/4+x√(1-x²)/4-arcsinx+C

Answer 2

∫ xarcsinx dx

= ∫ arcsinx d(x²/2)

= (1/2)x²arcsinx - (1/2)∫ x²/√(1 - x²) dx,x = sinz

= (1/2)x²arcsinx - (1/2)∫ sin²z/|cosz| * (cosz dz)

= (1/2)x²arcsinx - (1/2)∫ (1 - cos2z)/2 dz

= (1/2)x²arcsinx - (1/4)(z - 1/2*sin2z) + C

= (1/2)x²arcsinx - (1/4)arcsinx + (1/4)x√(1 - x²) + C

扩展资料：

分部积分：

(uv)'=u'v+uv'

得：u'v=(uv)'-uv'

两边积分得：∫ u'v dx=∫ (uv)' dx - ∫ uv' dx

即：∫ u'v dx = uv - ∫ uv' d,这就是分部积分公式

也可简写为：∫ v du = uv - ∫ u dv

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