Question

Помогите решить интеграл, пример: $\int\limits {\frac{dx}{4sinx-cosx+3} }$

Answer 1

Ответ:

Пошаговое объяснение:

$\int {\frac{dx}{4\sin x-\cos x+3} } \,=\begin{Vmatrix} t=\rm tg \frac{x}{2}, \; \sin x=\frac{2t}{1+t^2}, \; \cos x=\frac{1-t^2}{1+t^2}\\x=2\rm arctg \; t, \; dx=\frac{2}{1+t^2}dt \end{Vmatrix} =\int {\frac{\frac{2}{1+t^2} }{4\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2} +3 } } \,dt=\\\int {\frac{\frac{2}{1+t^2} }{\frac{8t-1+t^2+3+3t^2}{1+t^2} } } \, dt =\int {\frac{2(1+t^2)}{(1+t^2)(4t^2+8t+2)} } \,dt=\int {\frac{dt}{2t^2+4t+1} } \,=\int {\frac{dt}{((t\sqrt{2})^2+2*t\sqrt{2}*\sqrt{2}+(\sqrt{2})^2)-1 } } \,=$

$=\int {\frac{dt}{(t\sqrt{2}+\sqrt{2})^2 -1^2 } } \,=\frac{1}{\sqrt{2} } \int {\frac{dt\sqrt{2} }{(t\sqrt{2}+\sqrt{2})^2 -1^2 } } \,=\frac{1}{\sqrt{2} } \int {\frac{d(t\sqrt{2}+\sqrt{2}) }{(t\sqrt{2}+\sqrt{2})^2 -1^2 } } \,=\\\frac{1}{\sqrt{2} }*\frac{1}{2}\ln |\frac{t\sqrt{2}+\sqrt{2}-1 }{t\sqrt{2}+\sqrt{2}+1 } | +C=\frac{1}{2\sqrt{2} }\ln |\frac{\sqrt{2}\rm tg\frac{x}{2} +\sqrt{2}-1 }{\sqrt{2}\rm tg\frac{x}{2} +\sqrt{2}+1 } | +C$