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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= 2 \cdot \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

= 2 \cdot \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

= 2 \cdot \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)

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

Simplify

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

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

Add a constant to the solution

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

\int 4 1 - x^{2} d x

(5 x - 2 x^{4})^{3}

(x^{- 2})^{^{'}}

\int_{2}^{3} \frac{1}{x} d x

\int (6 x^{2} - 3 x + 3) d x

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