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
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?
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