How would I prove these are equal with Calculus?

Question

How would I prove that the circumference of an ellipse with equation x^2 + 2y^2 = 2 is the same as the arc length of y=sin(x) on the interval [0,2pi]?

Are they even equal?

Answer 1

Circumference of ellipse x^2 + 2y^2 = 2
Let x = √2 sin θ, y = cos θ, then x^2 + 2y^2 = 2 sin^2 θ + 2 cos^2 θ = 2, so x and y trace out the given ellipse as θ varies from 0 to 2π.
dx/dθ = √2 cos θ and dy/dθ = -sin θ, so the circumference is given by
∫(0 to 2π) √((dx/dθ)^2 + (dy/dθ)^2)
= ∫(0 to 2π) √((√2 cos θ)^2 + (-sin θ)^2)
= ∫(0 to 2π) √(2 cos^2 θ + sin^2 θ)
= ∫(0 to 2π) √(1 + cos^2 θ)
= arc length of y = sin(x) on [0, 2π]

Answer 2

The spinoff of tanhx is truly one million/(coshx)^2. right this is the info why. you may truly coach that the spinoff of cosh is sinh and the spinoff of sinh is cosh. as quickly as you do this, tanhx=sinhx/coshx. quotient rule is spinoff of a/b=(bda-adb)/b^2. so the spinoff of tanhx=(sinhx^2-coshx^2) / (coshx)^2 after algebra, it equals (sinhx+coshx) (sinhx-coshx) / (coshx)^2. changing those into the e^x format makes truly some the e^x and e^-x cancel out, leaving you with (e^x * e^-x) / (coshx)^2 = one million / (coshx)^2 :)