What is the integral of 2cos(4x)cos(9x) ?

The integral of 2cos(4x)cos(9x) is 1/5 sin(5x)+1/13 sin(13x)+C

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integral of 2cos(4x)cos(9x)

\int 2 cos (4 x) cos (9 x) d x

\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x) + C

Solution steps

\int 2 cos (4 x) cos (9 x) d x

Take the constant out:

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

= 2 \cdot \int cos (4 x) cos (9 x) d x

Rewrite using trig identities

= 2 \cdot \int \frac{1}{2} (cos (13 x) + cos (- 5 x)) d x

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int cos (13 x) + cos (- 5 x) d x

Simplify

cos (13 x) + cos (- 5 x) : cos (5 x) + cos (13 x)

= 2 \cdot \frac{1}{2} \cdot \int cos (5 x) + cos (13 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

= 2 \cdot \frac{1}{2} (\int cos (5 x) d x + \int cos (13 x) d x)

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

\int cos (13 x) d x = \frac{1}{13} sin (13 x)

= 2 \cdot \frac{1}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x))

Simplify

2 \cdot \frac{1}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x)) : \frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x)

= \frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x)

Add a constant to the solution

= \frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x) + C

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What is the integral of 2cos(4x)cos(9x) ?
The integral of 2cos(4x)cos(9x) is 1/5 sin(5x)+1/13 sin(13x)+C