Oblicz całkę \(\int \sin^2\!x\ dx\).
\[\begin{split} \int \sin^2\!x\ dx &=\int \frac{1-\cos 2x}{2}dx=\\[6pt] &=\frac{1}{2}\int1-\cos 2x\ dx=\\[6pt] &=\frac{1}{2}\int dx-\frac{1}{2}\int \cos 2x\ dx=\\[6pt] &=\frac{1}{2}x-\frac{1}{2} \frac{\sin 2x}{2}=\\[6pt] &=\frac{1}{2}x- \frac{\sin 2x}{4}+C=\\[6pt] &=\frac{x}{2}- \frac{2\sin x\cos x}{4}+C=\\[6pt] &=\frac{x-\sin x\cos x}{2}+C \end{split}\]