What is the integral of cos(2x+9)cos(2x-9) ?

The integral of cos(2x+9)cos(2x-9) is 1/2 (1/4 sin(4x)+cos(18)x)+C

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integral of cos(2x+9)cos(2x-9)

\int cos (2 x + 9) cos (2 x - 9) d x

\frac{1}{2} (\frac{1}{4} sin (4 x) + cos (18) x) + C

Solution steps

\int cos (2 x + 9) cos (2 x - 9) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (4 x) + cos (18)) 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 (4 x) + cos (18) 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 (18) d x)

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

\int cos (18) d x = cos (18) x

= \frac{1}{2} (\frac{1}{4} sin (4 x) + cos (18) x)

Add a constant to the solution

= \frac{1}{2} (\frac{1}{4} sin (4 x) + cos (18) x) + C

\int \frac{4}{x ^{2} - 2 x - 3} d x

\int e^{a t} sin (b t) d t

\int \frac{1}{x 49 + 36 x ^{2}} d x

\int \frac{4 x}{( x - 1 ) ( x + 1 ) ^{2}} d x

1 + sin^{2} (x)

What is the integral of cos(2x+9)cos(2x-9) ?
The integral of cos(2x+9)cos(2x-9) is 1/2 (1/4 sin(4x)+cos(18)x)+C