cos^2x/2的不定积分？

Question

高数不定积分

Answer 1

使用倍角公式
cos(x)^2/2=(1+cos2x)/4，然后积分得出(x+sin2x/2)/4+C

Answer 2

∫ cos2x / (cos²x * sin²x) dx
= ∫ cos2x / (cosx*sinx)² dx
= ∫ cos2x / (1/2 * 2sinx*cosx)² dx
= ∫ cos2x / [(1/2)² * (sin2x)²] dx
= ∫ cos2x / [(1/4) * sin²2x] dx
= 4∫ cos2x / sin²2x dx
= (4/2)∫ cos2x / sin²2x d(2x)
= 2∫ d(sin2x) / (sin2x)²
= -2/sin2x + C
= -2csc2x + C本回答被网友采纳