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

The integral from 0 to t of sin^2(wt) is 1/2 t-1/(4w)sin(2wt)

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

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

\frac{1}{2} t - \frac{1}{4 w} sin (2 wt)

Solution steps

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

Rewrite using trig identities

= \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

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

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

Simplify

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

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

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What is the integral from 0 to t of sin^2(wt) ?
The integral from 0 to t of sin^2(wt) is 1/2 t-1/(4w)sin(2wt)