Question

求积分，过程详细一点。

Answer 1

(4√2)[(1/3)cos³(Θ/2)-cos(Θ/2)]+C
解：
∫(1-cosΘ)^(3/2)dΘ
=∫[2sin²(Θ/2)]^(3/2)dΘ
=∫2^(3/2)sin³(Θ/2)dΘ
=(2√2)∫sin²(Θ/2)sin(Θ/2)dΘ
=(2√2)∫[1-cos²(Θ/2)]sin(Θ/2)dΘ
=(2√2)(-2)∫[1-cos²(Θ/2)]dcos(Θ/2)
=(4√2)∫[cos²(Θ/2)-1]dcos(Θ/2)
=(4√2)[(1/3)cos³(Θ/2)-cos(Θ/2)]+C

Answer 2

cosx = 1- 2[sin(x/2)]^2

∫( 1-cosx)^(3/2) dx
=2^(3/2) ∫ [sin(x/2)]^3 dx
=-2^(5/2) ∫ [sin(x/2)]^2 d cos(x/2)
=-2^(5/2) ∫ (1-[cos(x/2)]^2) d cos(x/2)
=-2^(5/2) .{ cos(x/2) - (1/3)[cos(x/2)]^3 } + C

Answer 3

这么简单，1-cosa=2cos2分之 a平方

Answer 4

将cosθ化为1-2sin^2θ/2

追答

这样就可以去根号去掉了

Answer 5

分