Another useful identity,sin t = 2v/(1+v²)

[MercuryRising]

a) find the area bounded by the polar curves r=3sinx, r=1+sinx

first i find the points of intersection, x=pi/6, and x=5pi/6
so to find the area i set up the integral
A= 1/2 integral from pi/6 to 5pi/6 [(3sinx)^2 - (1+sinx)^2]dx

using trig identities, it simplifies to

1/2 the integral from pi/6 to 5pi/6 [3 -4cos2x - 2sinx] dx

integrate and i get 1/2[3x - 2sin2x + 2cosx] from pi/6 to 5pi/6

which the sqrt 3's cancel out so i get (1/2)[ 5pi/2 - pi/2 ] = pi

i think this is correct but my answer is not on any of the answer choices
can someone please check what i did wrong?



b) find the length of the curve from t=0 to t=2pi given the curve has parametric equation
x= t- sint
y= 1 - cost

since the arc length ds is integral of sqrt[ (dx/dt)^2 + (dy/dt)^2]

dx/dt = 1 - cost
dy/dx = sint

thus ds = integral from 0 to 2 pi sqrt[(1-cost)^2 + (sint)^2]

which simplifies to integral from 0 to 2pi sqrt[ 2-2cost]
i couldn't integrate sqrt[ 2-2cost], or did i do something wrong in the steps prior to integration?

thanks

 

[Fermat]

a) I don't see anything wrong with your working or result. I get the same.

b) You're correct up to where you last got to. The integral is doable.
Use the substutution: v = tan(t/2) and it will come out.

 

[MercuryRising]

are you suggesting trig substitutions?
i thought you can only do that when the integral is in the form A^2 + V^2

 

[Fermat]

You can use trig substitutions at most anytime.

v = tan(t/2)
dv = (1/2)sec²(t/2) dt
2 dv = (1+v²) dt
dt = 2/(1+v²) dv
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Also,

cos t = (1- v²)/(1+v²)
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