求下列不定积分.求详解（1）∫[(cos2x)／(cosx﹢sinx)]dx;(2)∫cot²xdx;(3)∫{(1+2x²)／[x²(1+x²)]}dx;(4)∫sin²(x／2)dx;(5)∫[(cos2x)／(sin²xcos²x)]dx;(6)∫[e^(x-4)]dx;求解

Answer 1

cos2x=cos²x-sin²x=(cosx+sinx)(cosx-sinx)
∫(cosx-sinx)dx=sinx+cosx+C
∫cot²xdx=∫(csc²x-1)dx=-cotx-x+C
x=tanθ,dx=sec²θdθ
∫(1+2tan²θ)sec²θdθ/tan²θsec²θ
=∫(cotθ+2)dθ=ln|sinθ|+2θ+C=ln|x/√(1+x²)|+2arctanx+C
∫sin²(x/2)dx=(1/2)∫(1-cosx)dx=(x-sinx)/2+C
∫cos2xdx/(sinxcosx)²=4∫cos2xdx/sin²2x
=2∫d(sin2x)/sin²2x=-2/sin2x+C
∫[e^(x-4)]dx=e^(x-4)+C