Задача №2362

\[ \int\frac{dx}{1+\tg{x}} =\left[ t=\tg{x};\; dx=\frac{dt}{t^2+1}. \right] =\int\frac{dt}{(t+1)\left(t^2+1\right)}=\\ =\int\left(\frac{1}{2}\cdot\frac{1}{t+1} -\frac{1}{2}\cdot\frac{t}{t^2+1} + \frac{1}{2}\cdot\frac{1}{t^2+1} \right)dt =\frac{1}{2}\cdot\left(\ln(t+1) -\frac{1}{2}\ln\left(t^2+1\right)+\arctg{t}\right)+C=\\ =\frac{1}{2}\cdot\left(x+\ln\frac{|1+\tg{x}|}{\sqrt{1+\tg^2{x}}}\right)+C =\frac{1}{2}\cdot\left(x+\ln|\sin{x}+\cos{x}| \right) + C \]