Question

find the most general antiderivative or indefinite integral.
int \frac { 1 + cos ( 3 t ) } { 2 } d t
int \frac { 1 + cos ( 3 t ) } { 2 } d t =

Explanation:

Step1: Split the integral

$$\int\frac{1 + \cos(3t)}{2}dt=\frac{1}{2}\int1dt+\frac{1}{2}\int\cos(3t)dt$$

Step2: Integrate term - by - term

For $\frac{1}{2}\int1dt$, using the formula $\int kdt=kt + C$ ($k = 1$ here), we get $\frac{1}{2}t$.
For $\frac{1}{2}\int\cos(3t)dt$, let $u = 3t$, then $du=3dt$ and $dt=\frac{1}{3}du$.
So $\frac{1}{2}\int\cos(3t)dt=\frac{1}{2}\times\frac{1}{3}\int\cos(u)du$.
Since $\int\cos(u)du=\sin(u)+C$, substituting back $u = 3t$, we have $\frac{1}{6}\sin(3t)$.

Answer:

$\frac{t}{2}+\frac{\sin(3t)}{6}+C$