Answer
$\int_C (2x-y)dx +(x+3y)dy = (\frac{5}{2}-6\pi)$
Work Step by Step
The path is given by,
$x = 4\sin t , y = 3\cos t$ from (0,3) to (4,0)
$\therefore ds = 4\cos t dt, dy = -3\sin t dt, 0\le t \le \pi/2$
$\int_C (2x-y)dx + (x+3y)dy \\
= \int_{t=0}^{\pi/2}[8\sin t -3\cos t].(4\cos t)dt - [4\sin t + 9 \cos t].3\sin t dt\\
= \int_{t=0}^{\pi/2} (5\sin t \cos t -12)dt\\
=[\frac{5}{2}\sin^2 t -12t]_0^{\pi/2}\\
=(\frac{5}{2} - 6\pi)$
