Вопрос в картинках...

Воспользуемся универсальной тригонометрической подстановкой.
$\displaystyle...=\bigg\{u=tg\bigg( \frac{x}{2} \bigg);\,\,\sin x= \frac{2u}{u^2+1} ;\,\, \cos x= \frac{1-u^2}{u^2+1} ;\,\,\,dx= \frac{2du}{u^2+1}\bigg\}=\\ \\ \\ =\int\limits { \frac{2du}{(u^2+1)( \frac{5(1-u^2)}{1+u^2} +3)} } \, =\int\limits { \frac{du}{4-u^2} } \, = \frac{1}{4} \ln\bigg| \frac{2+u}{2-u} \bigg|+C =\\ \\ \\ =\boxed{ \frac{1}{4} \ln\bigg| \frac{2+ tg \frac{x}{2} }{2-tg \frac{x}{2} } \bigg|+C}$