I know this looks like a stupid question, but I want to know what I'm doing wrong here.
In the integral, the function itself is a scalar that does not depend on coordinates. The differential length element of the loop is the vector. I used the $R \cos{\theta}\hat{i} + R \sin{\theta}\hat{j}$ to define the circular loop, took its derivative for its differential length vector, giving $(-R \sin{\theta}\hat{i} + R \cos{\theta}\hat{j})d\theta$. Then integrated the function from $\theta = 0$ to $\theta = 2\pi$. The result was $0$, obviously.
I know for a fact that the integrated result should not be zero, since it represents a potential field from a current carrying conductor loop, measured at a distance away from the loop. The function is an AC current, i.e. $I_o\sin{2\pi ft}$.
Ideally, I want to know how to exactly integrate over a loop. What am I missing to get the right integral? I tried replacing $\cos{\theta}$ and $\sin{\theta}$ with actual ratio of triangle sides, but the value $\sqrt{R^2 - x^2}$ gives problems because of sign duality.
Edit: The actual integral I want to perform is: $$\int_{\theta = 0}^{2\pi}{I_o\sin{(2\pi ft)}(-R \sin{\theta}\hat{i} + R \cos{\theta}\hat{j})d\theta}.$$ This is as described earlier in the question.