Question

интеграл (1+tgx)/sin2x

Answer 1

$intfrac{1+tgx}{sin2x}dx=intfrac{(1+t)(1+t^2)}{2t(1+t^2)}=frac{1}{2}int(frac{1}{t}+1)dt=frac{1}{2}(ln|t|+t)=\=frac{1}{2}(ln|tgx|+tgx)+C\tgx=t\x=arctgt\dx=frac{dt}{1+t^2}\sin2x=frac{2t}{1+t^2}$