What is the integral of cos(6x)cos(3x) ?

The integral of cos(6x)cos(3x) is 1/2 (1/9 sin(9x)+1/3 sin(3x))+C

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integral of cos(6x)cos(3x)

\int cos (6 x) cos (3 x) d x

\frac{1}{2} (\frac{1}{9} sin (9 x) + \frac{1}{3} sin (3 x)) + C

Solution steps

\int cos (6 x) cos (3 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (9 x) + cos (3 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 (9 x) + cos (3 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 (9 x) d x + \int cos (3 x) d x)

\int cos (9 x) d x = \frac{1}{9} sin (9 x)

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

= \frac{1}{2} (\frac{1}{9} sin (9 x) + \frac{1}{3} sin (3 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{9} sin (9 x) + \frac{1}{3} sin (3 x)) + C

\int x \cdot \frac{1}{1} d x

\frac{\partial}{\partial x} (2 x y + 1)

\int x sec^{2} (d) x d x

\frac{d y}{d x}, y^{3} = x

\int e^{- 3 x} x d x

What is the integral of cos(6x)cos(3x) ?
The integral of cos(6x)cos(3x) is 1/2 (1/9 sin(9x)+1/3 sin(3x))+C