Compute the line integral $\int_q xydx +(x^2+y^2)dy$, where q is the first part in the first quadrant of a counterclockwise oriented circle $x^2+y^2=1$.
When parameterized I get $f(x,y) = (\cos(t), \sin(t))$ and taking the first derivative of $f(x,y)$ I get $dx = -\sin(t), dy = cos(t)$ because we're in the first quadrant we have $0 \le t \le \frac{\pi}{2}$
Plugging this all in to the integral:
$$\int_0^{\frac{\pi}{2}}-\cos(t)\sin^2(t)dt + \int_0^{\frac{\pi}{2}}(\cos^2(t)+\sin^2(t))\cos(t) dt$$
Have I approached the integral coorectly? If otherwise, what do I need to do to improve my solution?
Yes what you have done is correct but you should write them together.
$ \displaystyle \int_q xy ~ dx +(x^2+y^2) ~ dy = \int_q \vec F \cdot dr$
Where the vector field $ \vec F = (xy, x^2 + y^2)$ and we need to find line integral over the path $q$, which is part of the unit circle in the first quadrant parametrized as $ r (t) = (\cos t, \sin t), 0 \leq t \leq \frac{\pi}{2}$
$r'(t) = ( - \sin t, \cos t)$
So the line integral is,
$ \displaystyle \int_0^{\pi/2}(\cos t \sin t, 1) \cdot (- \sin t, \cos t) ~ dt$
$ \displaystyle = \int_0^{\pi/2} \cos t (1 - \sin^2 t) ~ dt$
$ \displaystyle = \int_0^1 (1 - u^2) ~ du ~ ~ $ (using substitution $u = \sin t$)