What is the integral of cos(wt-Bz) ?

The integral of cos(wt-Bz) is 1/w sin(wt-Bz)+C

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integral of cos(wt-Bz)

\int cos (wt - B z) d t

\frac{1}{w} sin (wt - B z) + C

Solution steps

\int cos (wt - B z) d t

Rewrite using trig identities

= \int cos (wt) cos (B z) + sin (wt) sin (B z) 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

= \int cos (wt) cos (B z) d t + \int sin (wt) sin (B z) d t

\int cos (wt) cos (B z) d t = cos (B z) \frac{1}{w} sin (wt)

\int sin (wt) sin (B z) d t = - \frac{1}{w} cos (wt) sin (B z)

= cos (B z) \frac{1}{w} sin (wt) - \frac{1}{w} cos (wt) sin (B z)

Simplify

cos (B z) \frac{1}{w} sin (wt) - \frac{1}{w} cos (wt) sin (B z) : \frac{1}{w} sin (wt - B z)

= \frac{1}{w} sin (wt - B z)

Add a constant to the solution

= \frac{1}{w} sin (wt - B z) + C

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What is the integral of cos(wt-Bz) ?
The integral of cos(wt-Bz) is 1/w sin(wt-Bz)+C