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

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

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integral of cos^2(wt)

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

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

Solution steps

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

Rewrite using trig identities

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

\int 1 d t = t

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

= \frac{1}{2} (t + \frac{1}{2 w} sin (2 wt))

Add a constant to the solution

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

x \to - 3 lim (\frac{x}{x} + \frac{1}{3})

\frac{d}{d x} (x ln ∣ cos (x) ∣)

\int \frac{sin ( 8 x )}{8} d x

\int \frac{13}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d y}{d x} = y e^{- x^{2}}, y (4) = 1

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