What is the integral of cos(2x)cos(7x) ?

The integral of cos(2x)cos(7x) is 1/2 (1/5 sin(5x)+1/9 sin(9x))+C

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integral of cos(2x)cos(7x)

\int cos (2 x) cos (7 x) d x

\frac{1}{2} (\frac{1}{5} sin (5 x) + \frac{1}{9} sin (9 x)) + C

Solution steps

\int cos (2 x) cos (7 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (9 x) + cos (- 5 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 (- 5 x) d x

Simplify

cos (9 x) + cos (- 5 x) : cos (5 x) + cos (9 x)

= \frac{1}{2} \cdot \int cos (5 x) + cos (9 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 (5 x) d x + \int cos (9 x) d x)

\int cos (5 x) d x = \frac{1}{5} sin (5 x)

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

= \frac{1}{2} (\frac{1}{5} sin (5 x) + \frac{1}{9} sin (9 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{5} sin (5 x) + \frac{1}{9} sin (9 x)) + C

(x, y) \to (0, 0) lim (\frac{sin ( x y )}{x ^{2} + y ^{2}})

\int \frac{x}{x ^{4} + 12 x ^{3} + 36 x ^{2}} d x

\int_{- 2}^{2} (32 - 8 y^{2}) d y

\frac{d}{d x} ((\frac{3 x - 4}{5 x + 2})^{- 5})

f (x) = 5 x^{2} cos (x) + 4 x

What is the integral of cos(2x)cos(7x) ?
The integral of cos(2x)cos(7x) is 1/2 (1/5 sin(5x)+1/9 sin(9x))+C