Question

ИНТЕГРАЛ DX/(SINX-COSX) СРОЧНОО

Answer 1

$\displaystyle\int\frac{dx}{sinx-cosx}=\int\frac{2}{1+t^2}*\frac{1+t^2}{t^2+2t-1}dt=2\int\frac{dt}{t^2+2t-1}=\\=2\int\frac{d(t+1)}{(t+1)^2-2}=\frac{1}{\sqrt2}ln|\frac{t+1-\sqrt2}{t+1+\sqrt2}|+C=\\=\frac{1}{\sqrt2}ln|\frac{tg\frac{x}{2}+1-\sqrt2}{tg\frac{x}{2}+1+\sqrt2}|+C\\\\\\t=tg\frac{x}{2};x=2arctgt;dx=\frac{2}{1+t^2}dt\\sinx=\frac{2t}{1+t^2};cosx=\frac{1-t^2}{1+t^2}\\sinx-cosx=\frac{t^2+2t-1}{1+t^2}$