What is the integral of (cos(3x)cos(2x))/3 ?

The integral of (cos(3x)cos(2x))/3 is 1/6 (1/5 sin(5x)+sin(x))+C

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integral of (cos(3x)cos(2x))/3

\int \frac{cos ( 3 x ) cos ( 2 x )}{3} d x

\frac{1}{6} (\frac{1}{5} sin (5 x) + sin (x)) + C

Solution steps

\int \frac{cos ( 3 x ) cos ( 2 x )}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int cos (3 x) cos (2 x) d x

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{1}{2} (cos (5 x) + cos (x)) d x

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int cos (5 x) + cos (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}{3} \cdot \frac{1}{2} (\int cos (5 x) d x + \int cos (x) d x)

\int cos (5 x) d x = \frac{1}{5} sin (5 x)

\int cos (x) d x = sin (x)

= \frac{1}{3} \cdot \frac{1}{2} (\frac{1}{5} sin (5 x) + sin (x))

Simplify

\frac{1}{3} \cdot \frac{1}{2} (\frac{1}{5} sin (5 x) + sin (x)) : \frac{1}{6} (\frac{1}{5} sin (5 x) + sin (x))

= \frac{1}{6} (\frac{1}{5} sin (5 x) + sin (x))

Add a constant to the solution

= \frac{1}{6} (\frac{1}{5} sin (5 x) + sin (x)) + C

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What is the integral of (cos(3x)cos(2x))/3 ?
The integral of (cos(3x)cos(2x))/3 is 1/6 (1/5 sin(5x)+sin(x))+C