Let's assume the area required is between x=p and x=q. This curve has open arms, which means that some areas will be infinite. This imposes constraints on the choice of p and q. p2≥a2 fulfils one constraint. Also q>p. Consider the integral a∫qydx, where y=x√(x2/a2-1)). This is the result of taking the square root of both sides and making y a function of x.
Let x=asecθ, so dx=asecθtanθdθ. When x=a, secθ=1⇒θ=0. When x=q, secθ=q/a, cosθ=a/q, θ=arccos(a/q), tanθ=√(q2/a2-1)=√(q2-a2)/a.
Let J=∫x√(x2/a2-1))dx=a∫secθtan2θdθ=a∫sec3θdθ-a∫secθdθ.
Let I=∫sec3θdθ=∫secθ(sec2dθ). Let u=secθ, du=secθtanθ, dv=sec2θdθ, v=tanθ.
I=uv-∫vdu=secθtanθ-∫secθtan2θdθ=secθtanθ-∫sec3θdθ+∫secθdθ,
I=secθtanθ-I+∫secθdθ;
secθ=secθ(secθ+tanθ)/(secθ+tanθ)=(sec2θ+secθtanθ)/(tanθ+secθ),
2I=secθtanθ+ln|tanθ+secθ|, I=½secθtanθ+½ln|tanθ+secθ|.
J=a(½secθtanθ+½ln|tanθ+secθ|)-aln|tanθ+secθ|=½a(secθtanθ-ln|tanθ+secθ|).
When θ=0, J0=0; θ=arccos(a/q), Jarccos(a/q)=½(q/a)√(q2-a2)-½aln((√(q2-a2)+q)/a).
The area J=Jarccos(a/q)-J0=½(q/a)√(q2-a2)-½aln((√(q2-a2)+q)/a) if integration is over the interval [a,q].
However, this is the area above the x-axis. There is an equal area below the x-axis so the total area over the same interval is (q/a)√(q2-a2)-aln((√(q2-a2)+q)/a).