[TEX]I = \int_{\frac{\pi}6}^{\frac{\pi}3}\frac{sin^3x.cos^2x}{sinx + cosx}dx [/TEX]
[TEX]I = \int_{\frac{\pi}6}^{\frac{\pi}3}\frac{tan^3x}{tanx+1}dx = \int_{\frac{\pi}6}^{\frac{\pi}3} (tan^2x + 1 - tanx)dx - \int_{\frac{\pi}6}^{\frac{\pi}3}\frac{cosx}{sinx+cosx}dx = I_1 + I_2[/TEX]
[TEX]I_1 = \int_{\frac{\pi}6}^{\frac{\pi}3}(tan^2x + 1 - tanx)dx = tanx \|_{\frac{\pi}6}^{\frac{\pi}3} + ln{| cosx |} \|_{\frac{\pi}{6}}^{\frac{\pi}3} = ...[/TEX]
[TEX]I_2 = \int_{\frac{\pi}6}^{\frac{\pi}3}\frac{cosx}{sinx+cosx}[/TEX]
Đặt [TEX]x = \frac{\pi}2 - t[/TEX]
Ta có :
[TEX]2I_2 = \int_{\frac{\pi}6}^{\frac{\pi}3}dx = \frac{\pi}6dx \Rightarrow I_2 = \frac{\pi}{12}[/TEX]
[TEX]\Rightarrow I = I_1 - I_2 = .... [/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn