What is the integral of sin(3x)cos(9x) ?

The integral of sin(3x)cos(9x) is 1/2 (-1/12 cos(12x)+1/6 cos(6x))+C

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integral of sin(3x)cos(9x)

\int sin (3 x) cos (9 x) d x

\frac{1}{2} (- \frac{1}{12} cos (12 x) + \frac{1}{6} cos (6 x)) + C

Solution steps

\int sin (3 x) cos (9 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 3 x + 9 x ) + sin ( 3 x - 9 x )}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (3 x + 9 x) + sin (3 x - 9 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

= \frac{1}{2} (\int sin (3 x + 9 x) d x + \int sin (3 x - 9 x) d x)

\int sin (3 x + 9 x) d x = - \frac{1}{12} cos (12 x)

\int sin (3 x - 9 x) d x = \frac{1}{6} cos (6 x)

= \frac{1}{2} (- \frac{1}{12} cos (12 x) + \frac{1}{6} cos (6 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{12} cos (12 x) + \frac{1}{6} cos (6 x)) + C

What is the integral of sin(3x)cos(9x) ?
The integral of sin(3x)cos(9x) is 1/2 (-1/12 cos(12x)+1/6 cos(6x))+C