Question

How do I integrate trig functions

Answer 1

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.