Jambmaths question:
Evaluate $\int{(\cos 4x+\sin 3x)dx}$
Option A:
$\sin 4x-\cos 3x+k$
Option B:
$\sin 4x+\cos 3x+k$
Option C:
$\tfrac{1}{4}\sin 4x-\tfrac{1}{3}\cos 3x+k$
Option D:
$\tfrac{1}{4}\sin 4x+\cos 3x+k$
Jamb Maths Solution:
Note the following
$\int{\sin ax}dx=-\frac{1}{a}\cos ax+k$
$\int{\cos axdx}=\frac{1}{a}\sin ax+k$
Therefore, $\int{(\cos 4x+\sin 3x)dx}=\frac{1}{4}\sin 4x-\frac{1}{3}\cos 3x+k$
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