Jambmaths question:
Find $\int{\cos 4xdx}$
Option A:
$-\tfrac{3}{4}\sin 4x+k$
Option B:
$\tfrac{1}{4}\sin 4x+k$
Option C:
$-\tfrac{1}{4}\sin 4x+k$
Option D:
$\tfrac{3}{4}\sin 4x+k$
Jamb Maths Solution:
$\begin{align} & I=\int{\cos 4xdx} \\ & \text{Let }u=4x;\frac{du}{dx}=4,dx=\frac{du}{4} \\ & I=\int{\cos u\frac{du}{4}=\frac{1}{4}}\cos udu=\frac{1}{4}\sin u+k \\ & I=\frac{1}{4}\sin 4x+k \\\end{align}$
Jamb Maths Topic:
Year of Exam: