Question

∫[π，0]xsinxdx=？

为什么等于派，
求详细解
还有一道简单的
（1）im sin(x^2-1)/x-1=?
x趋于1
怎用的无穷小变量啊
（2）∫x/√1+x^2dx=?

Answer 1

令u=π/2 - x
则u的取值范围是【-π/2 ，π/2】
原积分=∫[-π/2 ，π/2] (π/2 - u)sin(π/2 - u) d(π/2 - u)
= -∫[-π/2 ，π/2] (π/2 - u) cos u du
= -π/2∫[-π/2 ，π/2] cos u du + ∫[-π/2 ，π/2] u cos u du
u cos u是奇函数，故 ∫[-π/2 ，π/2] u cos u du=0
则原积分 = -π/2∫[-π/2 ，π/2] cos u du +0
= π/2∫[π/2 ，-π/2] cos u du
= (π/2) sin u| [π/2 ，-π/2]
= (π/2) ×[1-(-1)]
=π

（1）lim sin(x^2-1)/x-1=
x趋于1时，x^2-1趋于0，则sin(x^2-1)趋于0，因此sin(x^2-1)～x^2-1
则lim sin(x^2-1)/x-1=lim (x^2-1)/(x-1) =lim x+1=2

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