What is the integral of sin(wx+pi/3)sin(wx) ?

The integral of sin(wx+pi/3)sin(wx) is 1/w ((sqrt(3))/4 sin^2(wx)+1/4 (wx-1/2 sin(2wx)))+C

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integral of sin(wx+pi/3)sin(wx)

\int sin (w x + \frac{π}{3}) sin (w x) d x

\frac{1}{w} (\frac{3}{4} sin^{2} (w x) + \frac{1}{4} (w x - \frac{1}{2} sin (2 w x))) + C

Solution steps

\int sin (w x + \frac{π}{3}) sin (w x) d x

Apply u-substitution

= \int \frac{sin ( u ) ( \frac{3}{2} cos ( u ) + \frac{1}{2} sin ( u ) )}{w} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{w} \cdot \int sin (u) (\frac{3}{2} cos (u) + \frac{1}{2} sin (u)) d u

Expand

sin (u) (\frac{3}{2} cos (u) + \frac{1}{2} sin (u)) : \frac{3}{2} sin (u) cos (u) + \frac{1}{2} sin^{2} (u)

= \frac{1}{w} \cdot \int \frac{3}{2} sin (u) cos (u) + \frac{1}{2} sin^{2} (u) d u

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}{w} (\int \frac{3}{2} sin (u) cos (u) d u + \int \frac{1}{2} sin^{2} (u) d u)

\int \frac{3}{2} sin (u) cos (u) d u = \frac{3}{4} sin^{2} (u)

\int \frac{1}{2} sin^{2} (u) d u = \frac{1}{4} (u - \frac{1}{2} sin (2 u))

= \frac{1}{w} (\frac{3}{4} sin^{2} (u) + \frac{1}{4} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = w x

= \frac{1}{w} (\frac{3}{4} sin^{2} (w x) + \frac{1}{4} (w x - \frac{1}{2} sin (2 w x)))

Add a constant to the solution

= \frac{1}{w} (\frac{3}{4} sin^{2} (w x) + \frac{1}{4} (w x - \frac{1}{2} sin (2 w x))) + C

What is the integral of sin(wx+pi/3)sin(wx) ?
The integral of sin(wx+pi/3)sin(wx) is 1/w ((sqrt(3))/4 sin^2(wx)+1/4 (wx-1/2 sin(2wx)))+C