求不定积分∫sin2xcosxdx

Question

如题所述

Answer 1

3。求不定积分∫sin2xcosxdx
解：原式=2∫sinxcos²xdx=-2∫cos²xd(cosx)=-2[(1/3)cos³x]+C=-(2/3)cos³x+C
4。求定积分【0，1】∫{[arcsin(√x)]/√[x(1-x)]}dx
解：令arcsin(√x)=u，则√x=sinu，x=sin²u，dx=2sinucosudu；x=0时u=0；x=1时u=π/2；
故原式=【0，π/2】∫2usinucosudu/√[sin²u(1-sin²u)]=【0，π/2】∫2usinucosudu/(sinucosu)=【0，π/2】2∫udu=2(u²/2)【0，π/2】=π²/4.

Answer 2

因为sin2x = 2sinxcosx;
∫sin2xcosxdx = ∫2sinxcosxcosxdx = -2∫cosx^2dcosx = -2/3∫cosx^3追问

第四题呢？？

追答

本回答被提问者采纳

Answer 3

∫sin2xcosxdx
=∫2cosxsinxcosxdx
=－2∫cosxcosxdcosx
=－(2/3)(cosx)^3十C