1.求lim(n→∞)(1+2^n+3^n)^1/n 2.∫[0,π][sin(x/2)]^6

来源：学生作业帮编辑：搜搜考试网作业帮分类：数学作业时间：2024/06/08 07:16:38

lim(n→∞) (1 + 2^n + 3^n)^(1/n)
= e^lim(n→∞) [ln(1 + 2^n + 3^n)]/n
= e^lim(n→∞) [(2^n)ln2 + (3^n)ln3]/(1 + 2^n + 3^n)
= e^lim(n→∞) [(2/3)^n·ln2 + ln3]/[1/3^n + (2/3)^n + 1]
= e^[(0·ln2 + ln3)/(0 + 0 + 1)]
= e^(ln3)
= 3
∫(0→π) sin⁶(x/2) dx、y = x/2、dy = dx/2
= 2∫(0→π/2) sin⁶y dy
= 2·5!/6!·π/2、Wallis公式
= 2·(5·3·1)/(6·4·2)·π/2
= 5π/16
sin⁶y = (sin²y)³ = [(1 - cos2y)/2]³
= (1/8)(1 - 3cos2y + 3cos²2y - cos³2y)
= 1/8 - (3/8)cos2y + (3/8)[(1 + cos4y)/2] - (1/8)cos²2y·cos2y
= 5/16 - (3/8)cos2y + (3/16)cos4y - (1/8)[(1 + cos4y)/2]·cos2y
= 5/16 - (3/8)cos2y + (3/16)cos4y - (1/16)(1 + cos4y)cos2y
= 5/16 - (3/8)cos2y + (3/16)cos4y - (1/16)cos2y - (1/16)cos4ycos2y
= 5/16 - (7/16)cos2y + (3/16)cos4y - (1/16)(1/2)(cos6y + cos2y)
= 5/16 - (15/32)cos2y + (3/16)cos4y - (1/32)cos6y
∫(0→π/2) sin⁶(x/2) dx
= 2∫(0→π/2) sin⁶y dy
= 2∫(0→π/2) [5/16 - (15/32)cos2y + (3/16)cos4y - (1/32)cos6y] dy
= 2·5/16·π/2 - (15/32)(1/2)sin2y + (3/16)(1/4)sin4y - (1/32)(1/6)siny |(0→π/2)
= 5π/16

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