Question

интеграл cos^2x*dx/sin^4x
интеграл dx/cos^4x*sin^2x

Answer 1

$int frac{cos^2xdx}{sin^4x}=int frac{cos^2x}{sin^2x}cdot frac{dx}{sin^2x}=int ctg^2xcdot frac{dx}{sin^2x}=[t=ctgx,; dt=-frac{dx}{sin^2x}]=\\=-int t^2cdot dt=-frac{t^3}{3}+C=-frac{ctg^3x}{3}+C\\\int frac{dx}{cos^4xcdot sin^2x}=int frac{frac{dx}{cos^6x}}{frac{cos^4xsin^2x}{cos^6x}}=int frac{(frac{1}{cos^2x})^2cdot frac{dx}{cos^2x}}{tg^2x}=\\=[t=tgx,dt=frac{dx}{cos^2x},1+tg^2x=frac{1}{cos^2x}]=int frac{(1+t^2)^2dt}{t^2}=int frac{1+2t^2+t^4}{t^2}dt=$

$=int (t^{-2}+2+t^2)dt=frac{t^{-1}}{-1}+2t+frac{t^3}{3}+C=-frac{1}{tgx}=2tgx+frac{1}{3}cdot tg^3x+C$