I=\int_{}^{}[TEX]1/(cosx)^4.(sinx)'dx[/TEX] =\int_{}^{}[TEX]1/(1-sin^2x)^2.(sinx)'.dx[/TEX]
đặt sinx=t \RightarrowI=\int_{}^{}[TEX]1/(1-t^2)^2dt[/TEX]=1/2.\int_{}^{}[TEX](1/(t-1)-1/(t+1))^2.dt[/TEX]
=1/2.\int_{}^{}[TEX](1/(t-1)^2-2./(t^2-1)+1/(t+1)^2)[/TEX]=1/2.(-1/(t-1)-ln ((t-1)/(t+1))-1/(t+1))+c