求不定积分 1:∫x^2(sin)^2dx 2:∫e^(-2x)cosxdx 3:∫ln{x+根号(x^2+1)}dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/13 06:20:45
![求不定积分 1:∫x^2(sin)^2dx 2:∫e^(-2x)cosxdx 3:∫ln{x+根号(x^2+1)}dx](/uploads/image/z/679641-33-1.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86+1%EF%BC%9A%E2%88%ABx%5E2%28sin%29%5E2dx+2%3A%E2%88%ABe%5E%EF%BC%88-2x%EF%BC%89cosxdx+3%3A%E2%88%ABln%7Bx%2B%E6%A0%B9%E5%8F%B7%EF%BC%88x%5E2%2B1%EF%BC%89%7Ddx)
求不定积分 1:∫x^2(sin)^2dx 2:∫e^(-2x)cosxdx 3:∫ln{x+根号(x^2+1)}dx
求不定积分 1:∫x^2(sin)^2dx 2:∫e^(-2x)cosxdx 3:∫ln{x+根号(x^2+1)}dx
求不定积分 1:∫x^2(sin)^2dx 2:∫e^(-2x)cosxdx 3:∫ln{x+根号(x^2+1)}dx
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