题目

定积分（0,1）x^3*根号下(2+x^2)dx
给个思路就行.

提问时间：2020-10-15

答案

做法一：凑微分.
∫(0→1) x³√(2 + x²) dx
= ∫(0→1) [(2 + x²) - 2]√(2 + x²) d(x²/2)
= (1/2)∫(0→1) [(2 + x²)^(3/2) - 2√(2 + x²)] d(2 + x²),凑微分是不同转变积分限的
= (1/2)[(2/5)(2 + x²)^(5/2) - (4/3)(2 + x²)^(3/2)]：[0→1]
= (1/2)[(2/5)3^(5/2) - (4/3)3^(3/2)] - (1/2)[(2/5)2^(5/2) - (4/3)2^(3/2)]
= (8√2)/15 - (√3)/5
做法二：第二换元法.
先令x = √2 tanz,dx = √2sec²z dz
当x = 0,z = 0
当x = 1,z = arctan(1/√2)
∫(0→1) x³√(2 + x²) dx
= ∫ (√2 tanz)³ * √(2 + 2tan²z) * √2sec²z dz、0 ≤ z ≤ arctan(1/√2)
= 4√2∫ tan³z * sec³z dz
= 4√2∫ tan²zsec²z * secztanz dz
= 4√2∫ (sec²z - 1)sec²z d(secz)
= 4√2∫ (sec⁴z - sec²z) d(secz)
= 4√2 * [(1/5)sec⁵z - (1/3)sec³z]
= 4√2 * [(1/5)(√3/√2)⁵ - (1/3)(√3/√2)³] - 4√2 * [(1/5) - (1/3)]
= (8√2)/15 - (√3)/5
tanz = 1/√2
secz = √(tan²z + 1) = √(1/2 + 1) = √3/√2