Equal trigonometric expressions? I was doing $\int \frac12 \sin(2\theta)~\mathrm{d}\theta$, and got all the way to the second to last step, $-\frac{1}{4}\cos(2\,\theta\,)$. However, I don't understand the next step, which converts that expression to $-\frac{1}{2}\cos^2\,\theta$. Shouldn't there be a sine squared as well?
Also, if I plot out the two, the graphs do indeed look different. What is happening here?
(In case it matters, the original integral I was doing was $\int \sin\theta\cos\theta~\mathrm{d}\theta$, which I subsequently converted to the first integral I linked to)
 A: 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 $\int -v\,dv$.
A: The double angle formula gives
$$\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1.$$
You can plug that in and subsume the extra explicit constant into the arbitrary $+C$. (It's this vertical translation that explains why the two graphs look different; they are still both primitives.)
A: If we apply $\cos 2\theta = \cos^2{\theta}-\sin^2{\theta}$  and use identity $\cos^2{\theta} +\sin^2{\theta}=1$ we get next solution:
$\frac{-1}{2}\cos^2{\theta} + \frac{1}{4} +C$  and Wolfram Alpha made substitution $C'=\frac{1}{4} +C$, which is valid since $\frac{1}{4}$ is constant,  so final solution is of the form
$\frac{-1}{2}\cos^2{\theta}+C$
