Question

How do I find the integral to (sin(2x)+cos(3x)) dx?

Answer 1

\[\int\limits_{}^{}(\sin(2x)+\cos(3x)) dx = \int\limits \sin(2x) dx + \int\limits \cos(3x) dx\]can yu solve from here ?

Answer 2

-cos(2x)+sin(3x)+C doesn't look right to me since the derivative of that is sin(2x)*2+cos(3x)*3 so what am I doing wrong?

Answer 3

\[\int\limits \cos(bx) = -bsin(bx)+C\]\[\int\limits \sin(ax) = acos(ax)+C\]
does that help ?

Answer 4

all you have to do is adjust for the "thing" that comes out from the chain rule

Answer 5

I think your formulas are wrong MathMind. Or am I wrong?
\[\int\limits \cos{bx} dx = -\frac{1}{b}\sin{bx} + C\]

Answer 6

my frac thing didn't work i guess. Victor is right, it 1/b and 1/a