What is the integral of cos^2(kz-wt) ?

The integral of cos^2(kz-wt) is t/2-(zk)/(2w)-1/(4w)sin(2(-wt+zk))+C

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integral of cos^2(kz-wt)

\int cos^{2} (k z - wt) d t

\frac{t}{2} - \frac{z k}{2 w} - \frac{1}{4 w} sin (2 (- wt + z k)) + C

Solution steps

\int cos^{2} (k z - wt) d t

Apply u-substitution

: \int - \frac{cos ^{2} ( u )}{w} d u

= \int - \frac{cos ^{2} ( u )}{w} d u

Take the constant out:

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

= - \frac{1}{w} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= - \frac{1}{w} \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= - \frac{1}{w} \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

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}{w} \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

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

= - \frac{1}{w} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = - wt + z k

= - \frac{1}{w} \cdot \frac{1}{2} (- wt + z k + \frac{1}{2} sin (2 (- wt + z k)))

Simplify

- \frac{1}{w} \cdot \frac{1}{2} (- wt + z k + \frac{1}{2} sin (2 (- wt + z k))) : \frac{t}{2} - \frac{z k}{2 w} - \frac{1}{4 w} sin (2 (- wt + z k))

= \frac{t}{2} - \frac{z k}{2 w} - \frac{1}{4 w} sin (2 (- wt + z k))

Add a constant to the solution

= \frac{t}{2} - \frac{z k}{2 w} - \frac{1}{4 w} sin (2 (- wt + z k)) + C

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What is the integral of cos^2(kz-wt) ?
The integral of cos^2(kz-wt) is t/2-(zk)/(2w)-1/(4w)sin(2(-wt+zk))+C