How do you solve ∫sin(3x+4)dx?

I got the -cos(3x+4) part, but do you have to integrate the 3x+4 too? Does chain rule apply?

Use u substitution.

Let u = 3x+4. Then du = 3dx.
Substitute into the integral.
= (1/3)∫sin(u)du
= -(1/3)cos(u)+C
= -(1/3)cos(3x+4)+C

To solve ∫sin(3x+4)dx, you can indeed use the chain rule. However, keep in mind that when integrating, the chain rule works in reverse.

First, let's rewrite the integral as:
∫sin(3x+4)dx = ∫(sin(3x)cos(4) + cos(3x)sin(4))dx

Now, let's integrate each term separately.
∫sin(3x)cos(4)dx can be computed using the method of substitution. Let u = 3x, then du = 3dx.
The integral becomes:
∫sin(u)cos(4) * (1/3)du = (1/3)cos(4)∫sin(u)du = -(1/3)cos(4)cos(u) + C = -(1/3)cos(4)cos(3x) + C

Next, let's integrate the second term, ∫cos(3x)sin(4)dx. Again, we will use the method of substitution. Let u = 3x, then du = 3dx.
The integral becomes:
∫cos(u)sin(4) * (1/3)du = (1/3)sin(4)∫cos(u)du = (1/3)sin(4)sin(u) + C = (1/3)sin(4)sin(3x) + C

Finally, summing up both integrals, we get the result:
∫sin(3x+4)dx = -(1/3)cos(4)cos(3x) + (1/3)sin(4)sin(3x) + C

Therefore, the final solution is -(1/3)cos(4)cos(3x) + (1/3)sin(4)sin(3x) + C.