What is the integral of cos(4x)cos(5x) ?

The integral of cos(4x)cos(5x) is 1/2 (sin(x)+1/9 sin(9x))+C

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integral of cos(4x)cos(5x)

\int cos (4 x) cos (5 x) d x

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

Solution steps

\int cos (4 x) cos (5 x) d x

Rewrite using trig identities

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

Simplify

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

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

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

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

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

Add a constant to the solution

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

x^{2} + 6 x - 3

49 y^{^{''}} + 42 y^{^{'}} + 9 y = 0, y (0) = a, y^{^{'}} (0) = 2

x = \frac{( y - 3 ) ^{2}}{3}, y = 9 - x

7 x y \frac{d y}{d x} = 2

\frac{d}{d x} (1 + 5 x)

What is the integral of cos(4x)cos(5x) ?
The integral of cos(4x)cos(5x) is 1/2 (sin(x)+1/9 sin(9x))+C