[laTEX]I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{sin x - cosx+1}{sinx + 2cosx + 3}dx = \frac{1}{5}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 - 3.(cosx-2sinx)-(sin x + 2cosx+3)}{sinx + 2cosx + 3}dx \\ \\ I = \frac{1}{5}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8}{sinx + 2cosx + 3}dx - \frac{1}{5}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{ 3.(cosx-2sinx)}{sinx + 2cosx + 3}dx - \frac{1}{5}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{sin x + 2cosx+3}{sinx + 2cosx + 3}dx \\ \\ I = \frac{1}{5}(I_1-I_2 -I_3)[/laTEX]
[laTEX]I_3 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{sin x + 2cosx+3}{sinx + 2cosx + 3}dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx = \pi \\ \\ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{ 3.(cosx-2sinx)}{sinx + 2cosx + 3}dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{ 3.d(sin x + 2cosx +3)}{sinx + 2cosx + 3}dx \\ \\ I_2 = 3.ln|sinx + 2cosx +3| \big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ \\ I_1 : t = tan(\frac{x}{2}) \\ \\ cosx = \frac{1-t^2}{1+t^2} \\ \\ sinx = \frac{2t}{1+t^2}[/laTEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
