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tính nguyên hàm:
5sinx+cosx/2sinx-3cosx+1
[laTEX]I = \int \frac{(5sin x +cosx)dx}{2sinx-3cosx+1} \\ \\ \frac{1}{13}\int \frac{7(2sinx-3cosx+1) + 17(2cosx+3sinx) - 7}{2sinx-3cosx+1}dx \\ \\ I = \frac{1}{13}( \int \frac 7dx + \int \frac{17(2cosx+3sinx)dx}{2sinx-3cosx+1} - \int \frac{7dx}{2sinx-3cosx+1}) \\ \\ I = \frac{1}{13}( I_1+17I_2 -7I_3) [/laTEX]
I_1 dễ làm rồi
[laTEX]I_2 = \int \frac{(2cosx+3sinx)dx}{2sinx-3cosx+1} = ln|2sinx-3cosx+1| + C[/laTEX]
[laTEX]I_3 = \int \frac{dx}{2sinx-3cosx+1}[/laTEX]
dùng công thức sau
[laTEX]sinx = \frac{2tan(\frac{x}{2})}{1+tan^2(\frac{x}{2})} \\ \\ cosx = \frac{1 -tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})} \\ \\ u = tan(\frac{x}{2})[/laTEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
