Question

найти интеграл cosx*dx/sin^2-6sinx+5

Answer 1

$int frac{cosx, dx}{sin^2x-6sinx+5}=[t=sinx,; dt=cosxdx]=int frac{dt}{t^2-6t+5}=int frac{dt}{(t-3)^2-4}=\\=[z=t-3,; dz=dt]=int frac{dz}{z^2-4}=frac{1}{4}ln|frac{z-2}{z+2}|+C=frac{1}{4}ln|frac{sinx-5}{sinx-1}|+C$