What is the integral of sin(9x)cos(4x) ?

The integral of sin(9x)cos(4x) is 1/2 (-1/13 cos(13x)-1/5 cos(5x))+C

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integral of sin(9x)cos(4x)

\int sin (9 x) cos (4 x) d x

\frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

\int sin (9 x) cos (4 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 9 x + 4 x ) + sin ( 9 x - 4 x )}{2} 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 sin (9 x + 4 x) + sin (9 x - 4 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (13 x) + sin (5 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 sin (13 x) d x + \int sin (5 x) d x)

\int sin (13 x) d x = - \frac{1}{13} cos (13 x)

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

= \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{5} cos (5 x)) + C

\int_{1}^{2} I n x d x

y^{^{''}} + 2 y^{^{'}} + 6 y = 0

y^{^{''}} + 25 y = 20 sin (5 t)

2 s

(2 x + y) d x - x d y = 0

What is the integral of sin(9x)cos(4x) ?
The integral of sin(9x)cos(4x) is 1/2 (-1/13 cos(13x)-1/5 cos(5x))+C