u=x
dv=dx/cos^23x
du=dx
v= ∫ dx/cos^23x=(1/3) ∫d([blue]3x[/blue])x/cos^2[blue]3x [/blue]=(1/3)tg3x
получаем:
=(x*(1/3)tg(3x))|_(0)^(π/9)- ∫_(0)^(π/9)(1/3)tg3x dx=
=(π/27)tg(π/3)-0 - (1/3) ∫ _(0)^(π/9) tg (3x) dx=
=(π/27)sqrt(3) - (1/3)*(1/3) ∫ _(0)^(π/9) tg (3x) d(3x)=
=(π/27)sqrt(3) - (1/9)*(ln|cos3x|)| _(0)^(π/9)=
=(π/27)sqrt(3) - (1/9)*(ln|cos(π/3)|)|+(1/9) cos0=...