cos的六次方的不定积分怎么求解!

题目

cos的六次方的不定积分怎么求解!
那它在0到2/π的定积分是多少？

答案

so easy let me teach you.
cos⁶x
= (cos²x)³
= [(1 + cos2x)/2]³
= (1/8)(1 + cos2x)³
= (1/8)(1 + 3cos2x + 3cos²2x + cos³2x)
= 1/8 + (3/8)cos2x + (3/8)cos²2x + (1/8)cos²2xcos2x
= 1/8 + (3/8)cos2x + (3/8)(1 + cos4x)/2 + (1/8)(1 + cos4x)/2 · cos2x
= 1/8 + (3/8)cos2x + 3/16 + (3/16)cos4x + (1/16)cos2x + (1/16)cos4xcos2x
= 5/16 + (7/16)cos2x + (3/16)cos4x + (1/16)(1/2)(cos6x + cos2x)
= 5/16 + (15/32)cos2x + (3/16)cos4x + (1/32)cos6x
∴∫ cos⁶x dx
= 5x/16 + (15/32)(1/2)sin2x + (3/16)(1/4)sin4x + (1/32)(1/6)sin6x + C
= 5x/16 + (15/64)sin2x + (3/64)sin4x + (1/192)sin6x + C
楼上那个方法用的对,但是算的不对.应该如下才是正确
∫ cos⁶x dx
= (1/8)∫ (1 + 3cos2x + 3cos²2x + cos³2x) dx
= (1/8)∫ dx + (3/8)∫ cos2x dx + (3/8)∫ cos²2x dx + (1/8)∫ cos²2x cos2x dx
= x/8 + (3/8)(1/2)sin2x + (3/8)(1/2)∫ (1 + cos4x) dx + (1/8)(1/2)∫ cos²2x dsin2x
= x/8 + (3/16)sin2x + (3/16)(x + 1/4 · sin4x) + (1/16)∫ (1 - sin²2x) dsin2x
= x/8 + (3/16)sin2x + 3x/16 + (3/64)sin4x + (1/16)[sin2x - (sin³2x)/3] + C
= 5x/16 + (1/4)sin2x + (3/64)sin4x - (1/48)sin³2x + C
错误的地方是第四步(1/16)∫ (1 - sin²2x) dsin2x = (1/16)(sin2x - (sin³2x)/3) ≠ (1/16)(x - (sin³2x)/3)
这个积分在0到π/2上可用特别公式.
∫(0→π/2) cos⁶x dx
= (6 - 1)!/6! · π/2
= 5/6 · 3/4 · 1/2 · π/2
= 5π/32
对于公式如∫(0→π/2) sinⁿ dx = ∫(0→π/2) cosⁿx dx,n > 1
当n是奇数时
= (n - 1)!/n! = (n - 1)/n · (n - 3)/(n - 2) · (n - 5)/(n - 4) · ... · 3/4 · 1/2
当n是偶数时
= (n - 1)!/n! · π/2 = (n - 1)/n · (n - 3)/(n - 2) · (n - 5)/(n - 4) · ... · 3/4 · 1/2 · π/2,多了个π/2