Question
Evaluate the following integrals:$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin^2\text{x}\text{ dx}$

Answer

$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin^2\text{x}\text{ dx}$$=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{1-\cos2\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}(1-\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi}{2}-0+\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$

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