Help! Ayron: Czy ktos moglby mi obliczyc dlugosc luku krzywej? Potrzebuje tego na jutro. Z gory dzieki x=cost + ln(tg(t/2)) y= sint pi/4<=t<=pi/2

Basia: L=∫_t1^t₂ √(^dx_dt)² + (^dy_dt)² dt

dx		1		1		1
	= −sint +		*		*		=
dt		tg(t/2)		cos2(t/2)		2

	cos(t/2)
−sint +		=
	2sin(t/2)cos2(t/2)

	1
−sint +		=
	2sin(t/2)cos(t/2)

	1
−sint +		=
	sint

−sin2t+1
	=
sint

cos2t
sint

dy
	= cost
dt

(^dx_dt)²+(^dy_dt)²=

cos4t
	+cos2t =
sin2t

cos4t+sin2t*cos2t
	=
sin2t

cos4t + (1−cos2t)*cos2t
	=
sin2t

cos2t
	=
sin2t

1−sin2t		1
	=		− 1
sin2t		sin2t

	1
L = ∫π/4π/2 (		−1) dt =
	sin2t

(−ctgt−t) |_π/4^π/2 = [−ctg(π/2)−π/2]−[−ctg(π/4)−π/4]= [−0−π/2]−[−1−π/4]= −π/2 +1 + π/4 = 1−π/4