[m] ∫^{\frac{π}{2}} _{\frac{π}{6}}ctgxdx= ∫^{\frac{π}{2}} _{\frac{π}{6}}\frac{cosx}{sinx}dx=[/m]
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[i]Замена переменной:[/i]
[m] u=sinx[/m]
[m] du=(sinx)`dx[/m]
[m] du=cosx dx[/m]
[m]\frac{π}{6} → \frac{1}{2}[/m] ; [m]\frac{π}{2} →1[/m]
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[m]= ∫^{\frac{π}{2}} _{\frac{π}{6}}\frac{du}{u}=(ln|u|)|^{1} _{\frac{1}{2}}=ln|1|-ln|\frac{1}{2}|=-ln\frac{1}{2}=ln2[/m]