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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Simplify

cos (12 x) + cos (- 6 x) : cos (6 x) + cos (12 x)

= \frac{1}{2} \cdot \int cos (6 x) + cos (12 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 cos (6 x) d x + \int cos (12 x) d x)

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

\int cos (12 x) d x = \frac{1}{12} sin (12 x)

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

Add a constant to the solution

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

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