What is the integral of (1-cos(2t))/2 ?

The integral of (1-cos(2t))/2 is 1/2 (t-1/2 sin(2t))+C

integral of (1-cos(2t))/2

\int \frac{1 - cos ( 2 t )}{2} d t

\frac{1}{2} (t - \frac{1}{2} sin (2 t)) + C

Solution steps

\int \frac{1 - cos ( 2 t )}{2} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 - cos (2 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d t - \int cos (2 t) d t)

\int 1 d t = t

\int cos (2 t) d t = \frac{1}{2} sin (2 t)

= \frac{1}{2} (t - \frac{1}{2} sin (2 t))

Add a constant to the solution

= \frac{1}{2} (t - \frac{1}{2} sin (2 t)) + C

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What is the integral of (1-cos(2t))/2 ?
The integral of (1-cos(2t))/2 is 1/2 (t-1/2 sin(2t))+C