What is the integral from 0 to t of cos^2(wt) ?

The integral from 0 to t of cos^2(wt) is 1/2 (t+(sin(2wt))/(2w))

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integral from 0 to t of cos^2(wt)

\int_{0}^{t} cos^{2} (wt) d t

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

Solution steps

\int_{0}^{t} cos^{2} (wt) d t

Rewrite using trig identities

= \int_{0}^{t} \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_{0}^{t} 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_{0}^{t} 1 d t + \int_{0}^{t} cos (2 wt) d t)

\int_{0}^{t} 1 d t = t

\int_{0}^{t} cos (2 wt) d t = \frac{sin ( 2 wt )}{2 w}

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

\int 3 a y^{2} d y

\frac{d}{d x} (\frac{x + 5}{x - 1})

x \to 0 lim (sin (\frac{π x}{2}))

\int \frac{3 x ^{2}}{x ^{2} - 5} d x

f (3) = \frac{1}{x} + (x^{2} - 1)^{3}

What is the integral from 0 to t of cos^2(wt) ?
The integral from 0 to t of cos^2(wt) is 1/2 (t+(sin(2wt))/(2w))