Question:
$\int_{0}^{\pi}\left(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}\right) d x$
Solution:
Let $I=\int_{0}^{\pi}\left(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}\right) d x$
$=-\int_{0}^{\pi}\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right) d x$
$=-\int_{0}^{\pi} \cos x d x$
$\int \cos x d x=\sin x=\mathrm{F}(x)$
By second fundamental theorem of calculus, we obtain
$I=\mathrm{F}(\pi)-\mathrm{F}(0)$
$=\sin \pi-\sin 0$
$=0$
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