What is the integral of cos(4x)cos(8x) ?

The integral of cos(4x)cos(8x) is 1/2 (1/4 sin(4x)+1/12 sin(12x))+C

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integral of cos(4x)cos(8x)

\int cos (4 x) cos (8 x) d x

\frac{1}{2} (\frac{1}{4} sin (4 x) + \frac{1}{12} sin (12 x)) + C

Solution steps

\int cos (4 x) cos (8 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (12 x) + cos (- 4 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 (12 x) + cos (- 4 x) d x

Simplify

cos (12 x) + cos (- 4 x) : cos (4 x) + cos (12 x)

= \frac{1}{2} \cdot \int cos (4 x) + cos (12 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 (4 x) d x + \int cos (12 x) d x)

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

\int cos (12 x) d x = \frac{1}{12} sin (12 x)

= \frac{1}{2} (\frac{1}{4} sin (4 x) + \frac{1}{12} sin (12 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{4} sin (4 x) + \frac{1}{12} sin (12 x)) + C

\frac{d}{d x} (\frac{6}{3 x ^{5}} + \frac{2}{5 x ^{2}} - \frac{7}{9 x})

(y^{2} - x^{2}) d x + (- 2 x y) d y = 0

y^{^{''}} - 4 y^{^{'}} - 12 y = 8 e^{- 2 x}

\frac{d}{d x} (x e^{x} cos (x))

\int (5 x^{2} + 2 x^{3}) d x

What is the integral of cos(4x)cos(8x) ?
The integral of cos(4x)cos(8x) is 1/2 (1/4 sin(4x)+1/12 sin(12x))+C