How do I integrate trig functions

Integrate sin (3x+1) with respect to x. Use the substitution method. Let u=3x+1 be y our substitution. Then du/dx = 3, and du = 3dx.

Then we have "integrate sin u with respect to u". Here's where things are just a bit tricky at first. You have "integrate sin (3x+1) with respect to x." If u = 3x+1, then du/dx = 3 and du=3dx. But we don't have 3dx; we have only dx. Manipulate du=3dx as follows: du/3 = dx.

Then, "integral of sin (3x+1) dx" = > "integral of sin u (du/3) "

and this is equivalent to (1/3) "integral of sin u du." We get:

(1/3) (-cos u) + c, or (-1/3) * cos u + c. Finally, subst. 3x+1 for du, obtaining

(-1/3) * cos (3x+1) + C (answer).

You should find the derivative of this as a check. If the result of your differentiation is sin (3x+1), your integral was correct. Otherwise, it's not.

Don't worry ... this material becomes much easier once you've gotten the knack of it.