[m]tg\frac{x}{2}=t[/m] ⇒ [m]\frac{x}{2}=atctgt[/m] ⇒[m] x=2arctgt[/m] ⇒[m]dx=\frac{2dt}{1+t^2}[/m]
[m]sinx=\frac{2t}{1+t^2}[/m]
[m]cosx=\frac{1-t^2}{1+t^2}[/m]
[m]2sinx-cosx+5=2\cdot \frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}+5=\frac{4t-(1-t^2)+5(1+t^2)}{1+t^2}=\frac{3t^2+2t+2}{1+t^2}[/m]
подставляем [m]dx=\frac{2dt}{1+t^2}[/m] ; [m]2sinx-cosx+5=\frac{3t^2+2t+2}{1+t^2}[/m] в данный интеграл и
получим интеграл от дроби:
[m]\int \frac{dx}{2sinx-cosx+5}=\int \frac{dt}{3t^2+2t+2}=\frac{1}{3}\int \frac{dt}{t^2+\frac{2}{3}t+\frac{2}{3}}=\frac{1}{3}\int \frac{dt}{(t+\frac{1}{3})^2+\frac{5}{9}}=[/m]
[m]=\frac{1}{3}\int \frac{d(t+\frac{1}{3})}{(t+\frac{1}{3})^2+\frac{5}{9}}=\frac{1}{3}\frac{1}{\sqrt{\frac{5}{9}}}\cdot arctg\frac{t+\frac{1}{3}}{\sqrt{\frac{5}{9}}}+C=[/m]
[m]=\frac{1}{\sqrt{5}}\cdot arctg\frac{3(t+\frac{1}{3})}{\sqrt{5}}+C=\frac{1}{\sqrt{5}}\cdot arctg\frac{3t+1}{\sqrt{5}}+C[/m]