I want to calculate the mass of the ellipse curve $\vec\gamma(t)=(acos(t),bsin(t))$ for $a>0,b>0$ and for $t \in[0,2\pi]$, where line density $\lambda(x_1,x_2)=Cx_1x_2$ for $C>0$.
After plugging it to a formula for the line integral of a scalar field I get sth like that:
$\displaystyle\int_{0}^{\pi \over 2}C*a*cos(t)*b*sin(t)*\sqrt{a^2*sin^2(t)+b^2*cos^2(t)}dt$
And here is my question. How can I get rid of this root or do sth else to simplify this integral?
As you talk about a ellipse, I suppose $a\neq b$. Using $\cos^2 (t)=1-\sin^2 (t)$ you can write the square part of integrand as
$$\sqrt{a^2\,\sin^2 (t)+b^2\,\cos^2 (t)}=\sqrt{b^2+(a^2-b^2)\sin^2 (t)}$$
Now taking account that $(b^2+(a^2-b^2)\sin^2 (t))'=2(a^2-b^2)\sin(t)\cos(t)$ you integral can be written
$$\frac{C\,a\,b}{2(a^2-b^2)}\int_0^{\pi/2}2(a^2-b^2)\sin(t)\cos(t)\sqrt{b^2+(a^2-b^2)\sin^2 (t)}dt=$$ $$\left.=\frac{C\,a\,b}{3(a^2-b^2)}(b^2+(a^2-b^2)\sin^2 (t))^{3/2}\right|_0^{\pi/2}=\frac{C\,a\,b}{3(a^2-b^2)}(a^3-b^3)$$