What is the integral of cos(6x)cos(8x) ?

The integral of cos(6x)cos(8x) is 1/2 (1/2 sin(2x)+1/14 sin(14x))+C

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integral of cos(6x)cos(8x)

\int cos (6 x) cos (8 x) d x

\frac{1}{2} (\frac{1}{2} sin (2 x) + \frac{1}{14} sin (14 x)) + C

Solution steps

\int cos (6 x) cos (8 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (14 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 (14 x) + cos (- 2 x) d x

Simplify

cos (14 x) + cos (- 2 x) : cos (2 x) + cos (14 x)

= \frac{1}{2} \cdot \int cos (2 x) + cos (14 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 (14 x) d x)

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

\int cos (14 x) d x = \frac{1}{14} sin (14 x)

= \frac{1}{2} (\frac{1}{2} sin (2 x) + \frac{1}{14} sin (14 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} sin (2 x) + \frac{1}{14} sin (14 x)) + C

x \to 0 lim (- x)

\int 40 x d x

\frac{d}{d x} (\frac{6 x ^{2} - x ^{4}}{9})

\frac{\partial}{\partial x} (2 z^{3} \cdot y sin (x))

What is the integral of cos(6x)cos(8x) ?
The integral of cos(6x)cos(8x) is 1/2 (1/2 sin(2x)+1/14 sin(14x))+C