Solution 1:  We set $u =\sin(x)$ and $dv = \sin(x)\, dx$, so applying integration by parts gives $$\int \sin^2(x) \,dx \overset{\fbox{$\,\,u\,=\,\sin(x)\quad\,\,\, v\,=\,-\cos(x)\\du\,=\,\cos(x) dx\,\,\, dv\,=\,\sin(x)\,dx$}\\}{=} -\sin(x)\cos(x) + \int \cos^2(x) \,dx.$$ But $\cos^2(x)=1-\sin^2(x)$, so $\int \cos^2(x) \,dx=\int(1-\sin^2 x)\,dx=\int\,dx-\int\sin^2 x\,dx$.  We substitute this in, and then add this last term to the left-hand side from above, getting
$$\begin{eqnarray}\int \sin^2(x)\, dx &=& - \sin(x)\cos(x) + \int dx - \int \sin^2(x) \,dx \cr 2 \int \sin^2(x)\, dx &=& - \sin(x)\cos(x)+x + C \cr \int \sin^2(x) \,dx &=& \frac{x - \sin(x)\cos(x)}{2} + c\end{eqnarray}$$

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Solution 2:  $$\int \sin^2 x\, dx=\int \frac{1-\cos(2x)}{2}\,dx=\frac{1}{2}\int(1-\cos(2x))\,dx=\frac{1}{2}\left(x-\frac{1}{2}\sin(2x)\right)+c=\frac{1}{2}\left(x-\frac{1}{2}2\sin x\cos x\right)+c$$ $$= \frac{x - \sin(x)\cos(x)}{2} + c.$$   Here, we used another trig identity that you should know:  $\sin(2x)=2\sin x\cos x$.

Remember: always check to see that you have the right antiderivative by differentiating it!