Jambmaths question:
Evaluate $\int{\sin 2xdx}$
Option A:
$\tfrac{1}{2}\cos 2x+k$
Option B:
$-\tfrac{1}{2}\cos 2x+k$
Option C:
$-\cos 2x+k$
Option D:
$\cos 2x+k$
Jamb Maths Solution:
$\begin{align} & y=\int{\sin 2xdx} \\ & \text{Using change of variable } \\ & \text{Let }u=2x,\text{ }\frac{du}{dx}=2,\text{ }dx=\frac{du}{2} \\ & y=\int{\sin u\frac{du}{2}}=\frac{1}{2}\int{\sin udu} \\ & y=-\frac{1}{2}\cos u+K=-\frac{1}{2}\cos 2x+K \\\end{align}$
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