Question
$\int\sin^2\frac{\text{x}}{2}\text{dx}$

Answer

Let I $=\int\sin^2\frac{\text{x}}{2}\text{dx}.$ Then,
$\text{I}=\frac{1}{2}\int2\sin^2\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int(1-\cos\text{x})\text{dx}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=\frac{1}{2}\int\text{dx}-\frac{1}{2}\int\cos\text{xdx}$
$=\frac{1}{2}\times\text{x}-\frac{1}{2}\times\sin\text{x}+\text{C}$
$=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$
$\therefore\text{I}=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$

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