Compute the line integral $$\frac25\int_Cxds, $$ where the curve $C$ is the part of the curve of intersection between the ellipctic sylinder $$\big(\frac{x}{15}\big)^2 + \big(\frac{y}{5}\big)^2 = 1$$ and the plane $x = 3z$ lying in the first octant. Use the integral $$\int 2\sqrt{1+x^2}dx = x\sqrt{x^2 +1} + ln\bigg(x + \sqrt{x^2 +1}\bigg) + C $$
So I got the parametrizations to be: $$ x = 15\cos{t}, ~y = 5\sin{t}, ~z = 5\cos{t}$$ and $t$ ranging from $0$ to $\pi/2$. But I'm not able to get the integral in the form $\sqrt{x^2+1}$ so that I can solve the integral. So far what I have as the integrand after some simplifying is $$ 25\sqrt{9(\cos^2{t})(9\sin^2{t}) + 9\cos^2{t}}$$
First calculate ${\rm d}s$ as
\begin{eqnarray} {\rm d}s &=& \left[\left(\frac{{\rm d}x}{{\rm d}t} \right)^2 + \left(\frac{{\rm d}y}{{\rm d}t} \right)^2+ \left(\frac{{\rm d}z}{{\rm d}t} \right)^2\right]^{1/2} {\rm d}t \\ &=& \left[\left(5\cdot 3\sin t \right)^2 + \left(5\cos t \right)^2+ \left(5\sin t \right)^2\right]^{1/2} {\rm d}t \\ &=& 5\left[\left(3\sin t \right)^2 + 1\right]^{1/2} {\rm d}t \end{eqnarray}
So that
\begin{eqnarray} \int_Cx~{\rm d}s &=& \int_0^{2\pi} [15\cos t] 5\left[\left(3\sin t \right)^2 + 1\right]^{1/2} {\rm d}t \\ &=& 25 \int_0^{2\pi} (3\cos t)\left[\left(3\sin t \right)^2 + 1\right]^{1/2}{\rm d}t \end{eqnarray}
Now call $u = 3\sin t$ so that
$$ {\rm d} u = 3\cos t ~{\rm d}t $$
and
$$ \int (3\cos t)\left[\left(3\sin t \right)^2 + 1\right]^{1/2} {\rm d}t= \int [x^2 + 1]^{1/2}{\rm d}x $$
Can you take it from here?