Question

数学积分，简单一道题求解

求这个：
∫cos⁡(lnx)dx
（可以用 ∫udv=uv-∫vdu 这个公式，不用也行）

Answer 1

ʃ cos(ln(x)) dx
= (1/2) ʃ cos(ln(x)) dx + (1/2) ʃ cos(ln(x)) dx
= (1/2) ʃ cos(ln(x)) dx + (1/2) (x cos(ln(x)) + ʃ x sin(ln(x))/x dx)
= (1/2) ʃ cos(ln(x)) dx + (1/2) (x cos(ln(x)) + ʃ sin(ln(x)) dx)
= (1/2) ʃ cos(ln(x)) dx + (1/2) (x cos(ln(x)) + x sin(ln(x)) - ʃ cos(ln(x)) dx)
= (1/2) ʃ cos(ln(x)) dx + (1/2) x cos(ln(x)) + (1/2) x sin(ln(x)) - (1/2) ʃ cos(ln(x)) dx
= (1/2) x cos(ln(x)) + (1/2) x sin(ln(x))