Đặt [TEX]u=x [/TEX]
[TEX]dv = \frac{dx}{sin^2x} [/TEX]
[TEX]\Rightarrow du = dx [/TEX]
[TEX]v = -cotgx [/TEX]
Khi đó ta có
[TEX] I = - x.cotgx\mid\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} +\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cotgx.dx = (- x.cotgx + ln sinx)\mid\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \frac{\pi}{4} - ln (sin (\frac{\pi}{4}))[/TEX]