分部积分法
令 (3-2x)^(1/3) = u, 则 3-2x = u^3, x = (3-u^3)/2, dx = (-3/2)u^2du
I = (-3/4)∫(3-u^3)u^2du/u = (-3/4)∫(3u-u^4)du
= (-3/4)(3u^2/2 - u^5/5) + C = (-3/4)u^2(3/2 - u^3/5) + C
= (-3/40)(9+4x)(3-2x)^(2/3) + C
令 x = sinu, 则 dx = cosudu
I = 2∫<0, 1/2> xarcsinxdx/√(1-x^2) = 2∫<0, π/6> usinucosudu/cosu
= 2∫<0, π/6> usinudu = -2∫<0, π/6> udcosu
= -2[ucosu]<0, π/6> + 2∫<0, π/6> cosudu
= -√3π/6 + 2[sinu]<0, π/6> = 1-√3π/6
不懂找百度
如图所示。
2分之一，
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