Question 41

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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