What is the integral of sin(9x)cos(5x) ?

The integral of sin(9x)cos(5x) is 1/2 (-1/14 cos(14x)-1/4 cos(4x))+C

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integral of sin(9x)cos(5x)

\int sin (9 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{14} cos (14 x) - \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (9 x) cos (5 x) d x

Use the following identity:

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

= \int \frac{sin ( 9 x + 5 x ) + sin ( 9 x - 5 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 + 5 x) + sin (9 x - 5 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (14 x) + sin (4 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 (14 x) d x + \int sin (4 x) d x)

\int sin (14 x) d x = - \frac{1}{14} cos (14 x)

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

= \frac{1}{2} (- \frac{1}{14} cos (14 x) - \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{14} cos (14 x) - \frac{1}{4} cos (4 x)) + C

x \to 0 lim (8 x cot (3 x))

n = 1 \sum \infty (\frac{n + 1}{n})^{n}

y^{^{''}} - 2 y^{^{'}} + y = x e^{- x}

\frac{\partial}{\partial z} (x^{a} y^{b} z^{c})

\frac{d y}{d x} = \frac{( y - 1 )}{x + 3}

What is the integral of sin(9x)cos(5x) ?
The integral of sin(9x)cos(5x) is 1/2 (-1/14 cos(14x)-1/4 cos(4x))+C