exercise:
The upper arc of the ellipse $x^{2} / a^{2}+y^{2} / b^{2}=1$, with semi-major axis $a$ and semi-minor axis $b$ is the curve $$ y=\frac{b}{a} \sqrt{a^{2}-x^{2}}, \quad-a<x<a . $$ It is often convenient to express properties of the ellipse in terms of $a$, and the eccentricity $e$, defined as $$ e=\sqrt{1-\frac{b^{2}}{a^{2}}} . $$ Given $c$ such that $0<c<a$, express the integral for the arc length of the ellipse, from $x=0$ to $x=c$, in terms of $a, e$ and $c$.
Attempt:
In the previous exercise we have defined $\arcsin\,c$ as integral for the arc length of the curve $y=\sqrt{1-x^2}$ from $x = 0$ to $x = c$, where $0 < c < 1$. $$\arcsin(c)=\int_0^c\sqrt\frac{1}{1-x^2}\,dx$$
We can write the ellipse equation solved for $$y=\sqrt{(a^2-x^2)(1-e^2)}$$ and following the arc length formula we can write the arc length of the ellipse, from $x = 0$ to $x = c$ $$\int_0^c{\sqrt{\frac{a^2-x^2e^2}{a^2-x^2}}}\,dx$$ but where is "$c$". I'm not sure this is what the author intended. I guess the result of the previous exercise should have been used here. However, I don't know how to consider it here.
Can anybody help me, please?
P.S. I am familiar with the formulas containing $\cos x$ and $\sin x$ but meanwhile we have not defined them I cant use it.