반각 공식 \displaystyle \sin^{2}x=\dfrac{1-\cos(2x)}{2}를 사용한다.

\displaystyle \int_{0}^{\pi} \frac{1-\cos(2x)}{2}dx = \dfrac{1}{2}\left[x - \dfrac{1}{2}\sin(2x)\right]_{0}^{\pi}

= \displaystyle \frac{1}{2}\left[(\pi - \frac{1}{2}\sin(2\pi)) - (0 - \frac{1}{2}\sin(0))\right]

\displaystyle = \frac{1}{2}[(\pi - 0) - 0] = \frac{\pi}{2}