∫ dx / (1 + 3 cos² x)
tgx=t ⇒ 1+tg2x=[m]\frac{1}{cos^2x}[/m] ⇒ cos2x=[m]\frac{1}{1+tg^2x}[/m]
cos2x=[m]\frac{1}{1+t^2}[/m]
x=arctgt
dx=[m]\frac{dt}{1+t^2}[/m]
[m]\int \frac{dx}{1+3cos^2x}=\int \frac{\frac{dt}{1+t^2}}{1+3\cdot\frac{1}{1+t^2}}=\int \frac{dt}{1+t^2+3}=\int \frac{dt}{t^2+4}=[/m]
[m]=\frac{1}{2}\cdot arctg\frac{t}{2}=\frac{1}{2}\cdot arctg\frac{tgx}{2}+C[/m]– о т в е т