Line integral along ellipse, but with only one variable in integrand

Question

I have to calculate this integral:

$$\int_Lyds$$ where L is a part of ellipse $$\begin{cases} x=2\cos(t) \\ y=3\sin(t) & \ \end{cases}$$ in first quadrant.

The problem is the integrand contains only $y$ instead of $xy$. That's why "$u$ substitution" doesn't work, since it gives $du\over dt$ equals multiplication of $\sin(t)$ and $\cos(t)$ times some constant, but I have only $3\sin(t)$ to use as a part of my $du$.

Answer 1

hint

By Pythagoras,

$$ds=\sqrt {dx^2+dy^2}=$$

$$\sqrt {4\sin^2 (t)+9\cos^2 (t)}dt=$$ $$\sqrt{4+5\cos^2 (t)}dt .$$

the integral becomes

$$3\int_0^{\frac {\pi}{2}}\sqrt{4+5\cos^2 (t)}\sin (t)dt=$$

$$6\int_0^1\sqrt {1+\frac {5u^2}{4}}du $$

Now put $u=\frac {2\sinh (v)}{\sqrt {5}} $ and finish it.

Answer 2

Hint:

remember that

$$ ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt $$