инт 2х*соsx^4/sinx^9dx=[u=2x,du=2dx,dv=cosx^4/sinx^9,v=интегралсоsx^4/sinx^9dx]

вычислим инт соsx^4/sinx^9dx=[tgx/2=t, sin x=2t/1+t^2 ,cos x =1-t^2/1+t^2, dx=2dt/1+t^2],

инт(1-t^2/1+t^2)^4:(2t/1+t^2)^9*2dt/1+t^2= интеграл [(1-t^2)^4(1+t^2)^4]/2^8t^9 =

=1/2^8 инт[91-4t^4-2t^8+4t^8+4t^12+t^16)/t^9]dt