Answer
$\int(\frac{1}{3}sin(x)-\frac{1}{4}cos(x)) dx= -\frac{1}{3}cos(x) -\frac{1}{4}sin(x) +C $
Work Step by Step
Rules to remember:
$sin(x)' = cos(x)$, $cos(x)' = -sin(x)$
$\int(\frac{1}{3}sin(x)-\frac{1}{4}cos(x)) dx= -\frac{1}{3}cos(x) -\frac{1}{4}sin(x) +C $
