Question

求下列不定积分.求详解（1）∫[(cos 2x)／(cos x﹢sin x)]dx; (2) ∫cot²xdx

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