Question

分部积分法求∫x^2sin^2xdx，∫In^2xdx要过程

Answer 1

∫x²sin²x dx
=(1/2)∫x²(1-cos2x) dx
=(1/2)∫x² dx - (1/2)∫x²cos2x dx
=(1/2)(1/3)x³ - (1/2)(1/2)∫x² d(sin2x)
=(1/6)x³ - (1/4)x²sin2x + (1/4)∫sin2x*2x dx
=(1/6)x³ - (1/4)x²sin2x - (1/2)(1/2)∫x d(cos2x)
=(1/6)x³ - (1/4)x²sin2x - (1/4)xcos2x + (1/4)∫cos2x dx
=(1/6)x³ - (1/4)x²sin2x - (1/4)xcos2x + (1/8)sin2x + C

∫(lnx)² dx
=x(lnx)² - ∫x*2lnx*1/x dx
=x(lnx)² - 2∫lnx dx
=x(lnx)² - 2xlnx + 2∫x*1/x dx
=x(lnx)² - 2xlnx + 2x + C

Answer 2

我将它当不定积分求算，楼主自己代区间计算∫x/(sinx)^2dx＝∫x(cscx)^2dx ＝－∫xd(cotx)用分部积分法 ＝－xcotx＋∫cotxdx ＝－xcotx＋∫(cosx/