Question

Выручайте! $\int\limits {\frac{dx}{5-4sinx+7cosx} } \,$

Answer 1

$\int \dfrac{dx}{5-4sinx+7cosx}=\Big[\; t=tg\dfrac{x}{2}\; \Big]=\int \dfrac{2\, dt}{(1+t^2)\cdot (5-4\cdot \frac{2t}{1+t^2}+7\cdot \frac{1-t^2}{1+t^2})}=\\\\\\=\int \dfrac{2\, dt}{5+5t^2-8t+7-7t^2}=\int \dfrac{2\, dt}{-2t^2-8t+12}=-\dfrac{2}{2}\int \dfrac{dt}{t^2+4t-6}=\\\\\\=-\int \dfrac{dt}{(t+2)^2-10}=-\int \dfrac{d(t+2)}{(t+2)^2-10}=$

$=-\dfrac{1}{2\sqrt{10}}\cdot ln\left |\dfrac{t+2-\sqrt{10}}{t+2+\sqrt{10}}\right |+C=-\dfrac{1}{2\sqrt{10}}\cdot ln\left |\dfrac{tg\frac{x}{2}+2-\sqrt{10}}{tg\frac{x}{2}+2+\sqrt{10}}\right |+C$

Answer 2

Решение смотрите во вложении