Задача №1939

\[ \int\frac{\arcsin\tg{x}}{\cos^2{x}}dx =[t=\tg{x}] =\int\arcsin{t}dt =\left[\begin{aligned} & u=\arcsin{t};\;du=\frac{dt}{\sqrt{1-t^2}};\\ & dv=dt;\;v=t. \end{aligned}\right]=\\ =t\arcsin{t}-\int\frac{tdt}{\sqrt{1-t^2}} =t\arcsin{t}+\frac{1}{2}\cdot\frac{\left(1-t^2\right)^{\frac{1}{2}}}{\frac{1}{2}}+C =\tg{x}\cdot\arcsin\left(\tg{x}\right)+\sqrt{1-\tg^2{x}}+C. \]