Proving equality of trig identities I was asked to integrate the following: 
(A) $\int \sin2x dx$
and 
(B) $2 \int \sin x \cos x dx$, 
Which I have calculated becomes 
$-(1/2)\cos2x + (c/2) $
and 
$\sin^2x + (c/2) $
respectively. 
However, since $\sin^2x = 2\sin x\cos x$, these answers ought to be equal, right? 
If so, how can I prove they are equal?
 A: The function $-\frac{1}{2}\cos 2x$ is not identically equal to the function $\sin^2 x$.
But $\cos 2x=2\cos^2 x-1=\cos^2 x-\sin^2 x=1-2\sin^2 x$. Thus $\sin^2 x=-\frac{1}{2}\cos 2x +\frac{1}{2}$: your two functions differ by a constant. This is taken care of by the arbitrary "constant of integration." 
The family of functions of the shape $\sin^2 x+C$ is the same as the family of functions of the shape $-\frac{1}{2}\cos 2x +D$.
A: Well, the family of antiderivatives of $\sin(2x)$ is given by
$$\int\sin(2x)\,dx=-\frac12\cos(2x)+C,$$
where $C$ ranges over all real constants. Similarly, the family of antiderivatives of $2\sin(x)\cos(x)$ is given by
$$2\int\sin(x)\cos(x)\,dx=\sin^2(x)+K,$$
where $K$ ranges over all real constants. Now, using the fact that $\cos(2x)=1-2\sin^2(x),$ we see that $$-\frac12\cos(2x)+-\frac12+\sin^2(x)+C=\sin^2(x)+\left(C-\frac12\right).$$ Consequently, if we pick a particular antiderivative of $\sin(2x),$ we see that it is a particular antiderivative of $2\sin(x)\cos(x),$ since $C$ is then a fixed constant, and then $K=C-\frac12$ is a constant. Similarly, we see that any particular antiderivative of $2\sin(x)\cos(x)$ is a particular antiderivative of $\sin(2x)$.
