Question

这两道题的过程，谢谢

Answer 1

(1)
∫ (1/x^2) [cos(1/x)]^2 dx
=(1/2)∫ (1/x^2) [1+ cos(2/x)] dx
=-(1/2)(1/x) + (1/2)∫ (1/x^2). cos(2/x) dx
=-(1/2)(1/x) - (1/4)∫ cos(2/x) d(2/x)
=-(1/2)(1/x) - (1/4)sin(2/x) + C
(2)
∫(cosx)^3 dx
=∫(cosx)^2 dsinx
=∫[ 1-(sinx)^2] dsinx
=sinx -(1/3)(sinx)^3 + C
(3)
cos(A-B)= cosA.cosB +sinA.sinB (1)
cos(A+B)= cosA.cosB -sinA.sinB (2)
(1)-(2)
sinA.sinB = (1/2)[ cos(A-B)-cos(A+B)]
A=3x, B=5x
sin3x.sin5x =(1/州首2)[ cos2x-cos8x]
∫sin3x.sin5x dx
=∫(1/2)[ cos2x-cos8x] dx
=(1/2)[ -(1/2)sin2x + (1/8)sin8x ] +C
=-(1/4)sin2x + (1/轿迹伍16)sin8x +C
(4)
let
x = (2/3)sinu
dx=(2/3)cosu du
∫dx/√(4-9x^2)
=∫(2/3)cosu du /[ 2cosu]
=(1/3)∫闭或 du
=(1/3)u + C
=(1/3)arcsin(3x/2) + C