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

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

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integral of sin^2(wx)

\int sin^{2} (w x) d x

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

Solution steps

\int sin^{2} (w x) d x

Rewrite using trig identities

= \int 1 - cos^{2} (w x) d x

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 x - \int cos^{2} (w x) d x

\int 1 d x = x

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

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

Simplify

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

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

Add a constant to the solution

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

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