Question

Проинтегрировать тригонометрическое выражение: dx/(sin2x+cos2x)

Answer 1

$int frac{dx}{sin2x+cos2x}=[, t=tgx,; sin2x=frac{2t}{1+t^2},cos2x=frac{1-t^2}{1+t^2},\\x=arctgt,; dx=frac{dt}{1+t^2}, ]=int frac{frac{dt}{1+t^2}}{frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}}=int frac{dt}{-(t^2-2t-1)}=\\=-int frac{dt}{(t-1)^2-2} =[, u=t-1,; t=u+1,; du=dt, ]=\\=-int frac{du}{u^2-2}=-frac{1}{2sqrt2}cdot lnleft |frac{u-sqrt2}{u+sqrt2}right |+C=-frac{1}{2sqrt2}cdot lnleft |frac{tgx-1-sqrt2}{tgx-1+sqrt2}right |+C$