Question

来个高手，帮忙解下这道题（高数）

Answer 1

1、∫ xcosx/(sinx)^3dx
=∫ x/(sinx)^3d(sinx)
=(-1/2)∫ xd(sinx)^(-2)
=(-1/2)x(sinx)^(-2)+1/2∫ (sinx)^(-2)dx
=(-1/2)x(sinx)^(-2)+1/2∫ (cscx)^2dx
=(-1/2)x/(sinx)^2-1/2cotx+C

2、令√x=t，则x=t^2，dx=2tdt
原式=∫ e^t*2tdt
=2∫ td(e^t)
=2te^t-2∫ e^tdt
=2√xe^√x-2e^√x+C

Answer 2

∫xcosx/sin^3xdx
=∫x/sin^3xdsinx
=-1/2*x/sin^2x+1/2∫1/sin^2xdx
=-1/2*x/sin^2x+1/2∫csc^2xdx
=-1/2*x/sin^2x-1/2cotx+C

∫e^√xdx
令√x=t,x=t^2,dx=2tdt
∫e^√xdx
=∫e^t*2tdt
=2te^t-2∫e^tdt
=2te^t-2e^t+C

Answer 3

1)=
-(2*x + sin(2*x))/(4*sin(x)^2)+C
2)=
2*e^(x^(1/2))*(x^(1/2) - 1)+C