Jambmaths question:
Evaluate $\int{\sin 3xdx}$
Option A:
$\tfrac{2}{3}\cos 3x+C$
Option B:
$\tfrac{1}{3}\cos 3x+C$
Option C:
$-\tfrac{1}{3}\cos 3x+C$
Option D:
$-\tfrac{2}{3}\cos 3x+C$
Jamb Maths Solution:
$\begin{align}& y=\int{\sin 3xdx} \\ & \text{Using change of variable method, let }u=3x \\ & y=\int{\sin udx} \\ & \frac{du}{dx}=3,\text{ }dx=\frac{du}{3} \\ & \therefore y=\int{\sin u\frac{du}{3}=\frac{1}{3}\int{\sin udu}} \\ & y=-\frac{1}{3}\cos u+C \\ & y=-\frac{1}{3}\cos 3x+C \\\end{align}$
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