Question

Интеграл. Проверьте правильно ли я решил​

Answer 1

Используем универсальную тригонометрическую подстановку$u=tg \, (\frac{x}{2}); \ \ du=\frac{1}{2}\cdot \frac{1}{\cos^2{(\frac{x}{2}})} \\ \\ \sin{x}=\frac{2u}{u^2+1}; \ \ \cos{x}=\frac{1-u^2}{u^2+1}; \ \ \ dx=\frac{2 \, du}{u^2+1} \\ \\ \int {\frac{2}{(u^2+1)\cdot (1-\frac{3\cdot (1-u^2)}{u^2+1})}} \, du =\int {\frac{2}{(u^2+1)-3\cdot (1-u^2)}} \, du =\int {\frac{2}{u^2+1-3+3u^2}} \, du =\int {\frac{2}{4u^2-2}} \, du = \\ \\ = \int {\frac{1}{2u^2-1}} \, du = -\int {\frac{1}{1-2u^2}} \, du = -\frac{1}{\sqrt{2}}\int {\frac{d(\sqrt{2}u)}{1-(\sqrt{2}u)^2}}$

$=\frac{1}{\sqrt{2}}\cdot (-\frac{1}{2\cdot 1})\cdot \ln{|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}|} + C=-\frac{1}{2\sqrt{2}}\cdot \ln{|\frac{1+\sqrt{2}tg \, (\frac{x}{2})}{1-\sqrt{2}tg \, (\frac{x}{2})}|} + C=-\frac{\sqrt{2}}{4} \ln{|\frac{1+\sqrt{2}tg \, (\frac{x}{2})}{1-\sqrt{2}tg \, (\frac{x}{2})}|} + C$