Question

интеграл (1+tgx)/sin2x

Answer 1

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