Question
$\int\sin^2\text{bx dx}$

Answer

$\int\sin^2\text{bx dx}$
$=\int\Big[\frac{1-\cos2\text{bx}}{2}\Big]\text{dx}$ $\Big[\therefore\sin^2\text{x}=\frac{1-\cos2\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1-\cos2\text{bx})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{bx}}{2\text{b}}\Big]+\text{C}$
$=\frac{\text{x}}{2}-\frac{\sin2\text{bx}}{4\text{b}}+\text{C}$

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