This problem comes from my school's problem set, and right now we're on the cusp of finishing Calc 2 and starting Calc 3 in a few weeks.
For a given curve with $x = f(t)$, and $y = g(t)$, surface area$$S=\int_a^bg(t)\sqrt{(f'(t))^2 + g'(t)^2}dt$$
A semi-ellipse is given such that $$(x,y) = (a\cos(t), b\sin(t))$$ from $0\le t\le\pi$.
Thus, plugging everything in, I got the integral of
$$\int_a^bbsin(t)\sqrt{(-a\sin(t))^2 + (b\cos(t))^2}dt $$
Simplifying:$$\int_0^\pi b\sin(t)\sqrt{a^2\sin^2(t) + b^2\cos^2(t)}dt $$
How do I solve this though? I tried using an integral calculator and I was only told this is a special elliptical integral.
However, my school tells us that we are able to prove that the integral equals $$2\pi b\left[ b +\frac{a^2}{c}\arcsin\frac{c}{a}\right]$$
where $$c = \sqrt{a^2 + b^2}$$
Can anyone help me solve this integral?
