Question

求大神解答下面两题

Answer 1

(4)
let
u=√x
2udu = dx
x=0, u=0
x=1, u=1

∫(0->1) dx/(2+√x)
=∫(0->1) 2udu/(2+u)
=2∫(0->1) [1- 2/(2+u)] du
=2[u- 2ln|2+u|]|(0->1)
=2[(1-2ln3)-(0-2ln2)]
=2(1-2ln3+2ln2)

(5)

∫(0->π/2) xcosx dx
=∫(0->π/2) xdsinx
=[x.sinx]|(0->π/2) -∫(0->π/2) sinx dx
=π/2 +[cosx]|(0->π/2)
=π/2 +1