求不定积分 1：∫x^2(sin)^2dx 2:∫e^（-2x）cosxdx 3:∫ln{x+根号（x^2+1）}dx

题目

求不定积分 1：∫x^2(sin)^2dx 2:∫e^（-2x）cosxdx 3:∫ln{x+根号（x^2+1）}dx

答案

充分应用公式：∫udv=u*v-∫vdu; ∫du=∫u'dx
1.：∫x^2(sinx)^2dx =∫x^2*(1-cos2x)/2dx = ∫x^2/2dx - 1/4* ∫x^2*cos2xd(2x)
对于∫x^2/2dx= x^3/6
对于 - 1/4*∫x^2*cos2xd(2x) = - 1/4* ∫x^2dsin(2x) = -(x^2*sin2x)/4 + ∫sin(2x)dx^2
∫sin(2x)dx^2 = ∫x*sin2xd2x = - ∫xdcos2x = -x*cos2x+ ∫cos2xdx=sin2x/2 -x*cos2x
所以：：∫x^2(sinx)^2dx = x^3/6 + sin2x/2 -x*cos2x -(x^2*sin2x)/4 + C
2.∫e^（-2x）cosxdx = ∫e^(-2x)dsinx = e^(-2x)*sinx - ∫sinxde^(-2x)
= e^(-2x)*sinx +2* ∫e^(-2x)*sinxdx
=e^(-2x)*sinx - 2*∫e^(-2x)*dcosx
=e^(-2x)*sinx - 2*[ e^(-2x)*cosx - ∫cosxde^(-2x) ]
=e^(-2x)*sinx -2*e^(-2x)*cosx - 2*∫e^（-2x）cosxdx
所以∫e^（-2x）cosxdx = e^(-2x)*sinx /3 -2*e^(-2x)*cosx/3
3.令1/根号(x^2+1) = cost (-π/2=< t