How do you evaluate the integral int 3sinx+4cosxdx?

$\int 3 \sin \left(x\right) + 4 \cos \left(x\right) \mathrm{dx} = 4 \sin \left(x\right) - 3 \cos \left(x\right) + C$

Explanation:

This integral is easier than it looks!

The first thing that we need to realize is that we can break the integral up over the addition:

$\int 3 \sin \left(x\right) + 4 \cos \left(x\right) \mathrm{dx} = \int 3 \sin \left(x\right) \mathrm{dx} + \int 4 \cos \left(x\right) \mathrm{dx}$

Then we can move the constants out of the integrals:

$\int 3 \sin \left(x\right) \mathrm{dx} + \int 4 \cos \left(x\right) \mathrm{dx} = 3 \int \sin \left(x\right) \mathrm{dx} + 4 \int \cos \left(x\right) \mathrm{dx}$

Now we can just sub in the antiderivatives for $\sin \left(x\right)$ and $\cos \left(x\right)$.
You should have these memorized, they come up a lot!

$\int \sin \left(x\right) = - \cos \left(x\right) + {c}_{1}$
$\int \cos \left(x\right) = \sin \left(x\right) + {c}_{2}$

$3 \int \sin \left(x\right) \mathrm{dx} + 4 \int \cos \left(x\right) \mathrm{dx} = 3 \left(- \cos \left(x\right) + {c}_{1}\right) + 4 \left(\sin \left(x\right) + {c}_{2}\right)$

$\int 3 \sin \left(x\right) + 4 \cos \left(x\right) \mathrm{dx} = 4 \sin \left(x\right) - 3 \cos \left(x\right) + C$
note: I combined the two constants of integration, ${c}_{1}$ and ${c}_{2}$ into one constant $C$ because it's easier to deal with