Answer
$\int(2cos(x) - 9sin(x)) dx = 2sin(x) +9cos(x) + C$
Work Step by Step
Rules to remember:
$\int sin(x)dx = -cos(x)$, $\int cos(x)dx = sin(x)$
$sin(x)' = cos(x)$, $cos(x)' = -sin(x)$
Anti-differentiation: $\int(2cos(x) - 9sin(x)) dx = 2sin(x) -(-9cos(x)) + C= 2sin(x) +9cos(x) + C$
Checking with differentiation:
$(2sin(x) +9cos(x) + C)'= 2cos(x) -9sin(x)$
