Recall the double-angle trigonometric identities $\cos(2\theta)=2\cos^2(\theta) -1=\cos^2(\theta)-\sin^2(\theta)=1-2\sin^2(\theta).$ (We only need the first equality, but I might as well mention the other two common forms.)
It follows that $-\frac{1}{4}\cos(2\theta)=-\frac{1}{2}\cos^2(\theta) +\frac{1}{4}.\qquad\text{(Equation 1)}$ So your two expressions are indeed not equal.
By Equation 1, $-\frac{1}{4}\cos(2\theta)$ and $-\frac{1}{2}\cos^2(\theta)$ differ by a constant. When we are finding antiderivatives (indefinite integrals) there is always an arbitrary constant of integration.
Thus the two answers $-\frac{1}{4}\cos(2\theta)+C$ and $-\frac{1}{2}\cos^2(\theta)+C$ are both correct.
Here is a more obvious example. The result $\int 2x\,dx=x^2+C$ is correct. The result $\int 2x\,dx=x^2-\pi^3+C$ is also correct.
Comment: If we had used the identity $\cos(2\theta)=1-2\sin^2(\theta)$, we would have found in the same way that $\frac{1}{2}\sin^2(\theta)+C$ is another perfectly correct answer! It is the one I like best, no minus signs. But overall, there does not appear to be any urgent reason to fiddle with the correct $-\frac{1}{4}\cos(2\theta)+C$ that you first obtained. (Without the $+C$, all the various answers would be incorrect.)
If the original integral you were doing was, as you indicate, $\int \sin(\theta)\cos(\theta)\,d\theta$, there are two natural approaches.
1.) Use the identity $\sin(2\theta) =2\sin\theta\cos\theta$ to express the integral as $\int \frac{1}{2}\sin(2\theta)\,d\theta$. That seems to be how you approached things, and it works quickly.
2.) Note that the derivative of $\sin\theta$ is $\cos\theta$. Make the substitution $u=\sin(\theta)$. Then $du=\cos(\theta)\,d\theta$. We find that $\int \sin(\theta)\cos(\theta)\,d\theta=\int u\,du=\frac{u^2}{2}+C$. Substitute back, to get $\frac{\sin^2(\theta)}{2}+C$. This answer can be turned into various shapes by using trigonometric identities.
Or else we can start by making the substitution $v=\cos\theta$. Then $dv=-\sin(\theta)\,d\theta$, and we end up with $\int -v\,dv$.