用换元法求不定积分,cosx^4,答案为1/32sinx+1/4sin2x+3/8x,求过程.

Question

用换元法求不定积分,cosx^4,答案为1/32sinx+1/4sin2x+3/8x, 求过程.

Answer 1

当n是奇数时,∫ (cosx)^n dx才可用换元法,不然只能用配角公式逐步拆解,这题的n是偶数
∫ cos^4x dx
= ∫ (cos²x)² dx
= ∫ [1/2*(1+cos2x)]² dx
= (1/4)∫ (1+2cos2x+cos²2x) dx
= (1/4)∫ dx + (1/4)∫ cos2x d(2x) + (1/4)∫ 1/2*(1+cos4x) dx,若要要换元法,这里可以用
= (1/4)x + (1/4)sin2x + (1/8)(x + 1/4*sin4x) + C
= (1/4+1/8)x + (1/4)sin2x + (1/32)sin4x + C
= (1/32)sin4x + (1/4)sin2x + (3/8)x + C