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

The integral of sin^2(wt) is 1/2 t-1/(4w)sin(2wt)+C

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integral of sin^2(wt)

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

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

Solution steps

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

Rewrite using trig identities

= \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

= \int 1 d t - \int cos^{2} (wt) d t

\int 1 d t = t

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

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

Simplify

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

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

Add a constant to the solution

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

arctan (4 x)

\frac{d}{d x} ((x^{- 2} + x)^{3} \cdot (4 x)^{- 2 x})

\frac{\partial}{\partial x} (l - c)

\int x (15 x + 4) d x

\frac{( 6 x ^{2} + 6 x + 4 )}{x}

What is the integral of sin^2(wt) ?
The integral of sin^2(wt) is 1/2 t-1/(4w)sin(2wt)+C