Question

高数不定积分计算～～

Answer 1

∫cos(lnx) dx
let
u= lnx
du = (1/x) dx
dx = e^u .du

∫cos(lnx) dx
=∫ cosu .e^u du
=∫ cosu de^u
=cosu. e^u + ∫ sinu .e^u du
=cosu. e^u + ∫ sinu de^u
=cosu. e^u + sinu .e^u -∫ cosu .e^u du
2∫ cosu .e^u du = (sinu +cosu)e^u
∫ cosu .e^u du = (1/2) (sinu +cosu)e^u +C

∫cos(lnx) dx
=(1/2) (sinu +cosu)e^u +C
=(1/2)x. (sin(lnx) +cos(lnx)) +C

(2)
let
u= tanx
du = (secx)^2 .dx
∫(secx)^2 /(2+(tanx)^2) dx
=∫ du/( 2+ u^2)
=(√2/2)arctan(u/√2 ) + C
=(√2/2)arctan(tanx/√2 ) + C