The integral can be calculated using integration by parts (remember that $∫u’(x)v(x)\,dx = u(x)v(x) - ∫ u(x)v’(x)\,dx$):

$$ ∫ \sin(x)\sin(x)\,dx = -\cos(x)\sin(x)-∫(-\cos(x))\cos(x)\,dx $$If we apply integration by parts to the rightmost expression again, we will get $∫\sin^2(x)dx = ∫\sin^2(x)dx$, which is not very useful. The trick is to rewrite the $\cos^2(x)$ in the second step as $1-\sin^2(x)$. Then we get

\begin{align*} ∫ \sin^2(x)\,dx &= -\cos(x)\sin(x)+∫(1-\sin^2(x))\,dx \\ &= -\cos(x)\sin(x)+x-∫\sin^2(x)\,dx \end{align*}Now, all we have to do is to transfer $∫\sin^2(x)\,dx$ from the right-hand side to the left-hand side of the equation:

$$ 2∫ \sin^2(x)\,dx = -\cos(x)\sin(x)+x + C $$ $$ ∫ \sin^2(x)\,dx = ÷{1}{2}(x-\cos(x)\sin(x))+C $$(Technically speaking, we should write $C/2$ rather than $C$, but it is still just an indefinite constant, so, as is customary, we rename $C/2$ to $C$ to get the result in the usual form.)