∫[π，0]xsinxdx=？

∫[π，0]xsinxdx=？

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 duu 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