Question

Решить по интеграллам пож.) s dx / 4-5sinx

Answer 1

Универсальная тригонометрическая подстановка.
$\displaystyle \int\limits { \frac{dx}{4-5\sin x} } \, =\bigg\{ tg \frac{x}{2}=t;\,\, \sin x= \frac{2t}{t^2+1} ;\cos x= \frac{1-t^2}{1+t^2};\\ \\ \\ dx= \frac{2dt}{t^2+1} \bigg\}=\int\limits { \frac{dt}{2t^2-5t+2} } \, =\int\limits { \frac{dt}{( \sqrt{2}t- \frac{5}{2 \sqrt{2} } )^2- \frac{9}{8} } } \, =\\ \\ \\ =\bigg\{ \sqrt{2}t- \frac{5}{2 \sqrt{2} } =u\bigg\}= \frac{1}{\sqrt{2}} \int\limits { \frac{du}{u^2- \frac{9}{8} } } \, = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{3} \ln| \frac{4u-3 \sqrt{2} }{4u+3\sqrt{2}}|+C=$
$\displaystyle= \frac{1}{3} \ln| \frac{4( \sqrt{2}t- \frac{5}{2 \sqrt{2} })-3 \sqrt{2} }{4( \sqrt{2}t- \frac{5}{2 \sqrt{2} })+3 \sqrt{2} } |+C,\,\,\, when\,\,\,\, t=tg\bigg( \frac{x}{2}\bigg)$