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

The integral of cos^2(wt-kx) is -(wt)/(2k)+x/2-1/(4k)sin(2(wt-kx))+C

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integral of cos^2(wt-kx)

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

- \frac{wt}{2 k} + \frac{x}{2} - \frac{1}{4 k} sin (2 (wt - k x)) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Substitute back

u = wt - k x

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

Simplify

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

= - \frac{wt}{2 k} + \frac{x}{2} - \frac{1}{4 k} sin (2 (wt - k x))

Add a constant to the solution

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

\int \frac{1}{x ^{2} - 4} d x

x \to 0 + lim (lo g_{2} (x))

\int ln (x^{2} - 1) d x

\frac{\partial}{\partial y} (x^{2} ln (3 y - 5))

\frac{x ^{2} - 1}{x ^{2} + 1}

What is the integral of cos^2(wt-kx) ?
The integral of cos^2(wt-kx) is -(wt)/(2k)+x/2-1/(4k)sin(2(wt-kx))+C