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

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

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

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

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

Solution steps

\int cos (4 x) cos (2 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

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 (2 x) d x)

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (a r (x) cos (x))

\int \frac{- x - 1}{x ^{2} + 3} d x

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\frac{d}{d x} (x^{2} e^{5 x})

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