\ANG{
Beweise mittels einer sehr ausf"uhrlichen Rechnung:
\[\int\sin^{2}x\,\RMd x=\frac{x}{2}-\frac{\sin x\cos x}{2}+C\]
} 
\LSG{
\begin{align*}
\left(\frac{x}{2}-\frac{\sin x\cos x}{2}+C\right)'&=
\frac12-\frac12\left(\cos x\cos x+\sin x(-\sin x)\right)=\\
&=\frac12\left[1-\left(\cos^{2} x-\sin^{2} x\right)\right]=
\frac12\left[1-\cos^{2} x+\sin^{2} x\right]=\\
&=\frac12\left[\sin^{2} x+\sin^{2} x\right]=\sin^{2}x\qquad\text{q.e.d.}
\end{align*}
}
