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

The integral of cos(2x)cos(4x) is 1/2 (1/2 sin(2x)+1/6 sin(6x))+C

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

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

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

Solution steps

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

Rewrite using trig identities

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

Simplify

cos (6 x) + cos (- 2 x) : cos (2 x) + cos (6 x)

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

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

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

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

Add a constant to the solution

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

What is the integral of cos(2x)cos(4x) ?
The integral of cos(2x)cos(4x) is 1/2 (1/2 sin(2x)+1/6 sin(6x))+C