Question

不定积分问题2

Answer 1

观察d(2sinx - cosx) = (2cosx + sinx) dx
分子可变为cosx = A(2cosx + sinx) + B(2sinx - cosx) + C的形式
cosx = (2A - B)cosx + (A + 2B)sinx + C
C = 0
A + 2B = 0、A = - 2B
2A - B = 1、2(- 2B) - B = 1、- 5B = 1、B = - 1/5
A = - 2(- 1/5) = 2/5
∴
∫ cosx/(2sinx - cosx) dx
= (2/5)∫ (2cosx + sinx)/(2sinx - cosx) dx + (- 1/5)∫ (2sinx - cosx)/(2sinx - cosx) dx
= (2/5)∫ d(2sinx - cosx)/(2sinx - cosx) - (1/5)∫ dx
= (2/5)ln|2sinx - cosx| - x/5 + C