What is the integral of sin(18x)cos(13x) ?

The integral of sin(18x)cos(13x) is 1/2 (-1/31 cos(31x)-1/5 cos(5x))+C

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integral of sin(18x)cos(13x)

\int sin (18 x) cos (13 x) d x

\frac{1}{2} (- \frac{1}{31} cos (31 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

\int sin (18 x) cos (13 x) d x

Use the following identity:

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

= \int \frac{sin ( 18 x + 13 x ) + sin ( 18 x - 13 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 (18 x + 13 x) + sin (18 x - 13 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (31 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 (31 x) d x + \int sin (5 x) d x)

\int sin (31 x) d x = - \frac{1}{31} cos (31 x)

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

= \frac{1}{2} (- \frac{1}{31} cos (31 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{31} cos (31 x) - \frac{1}{5} cos (5 x)) + C

\int x^{2} \cdot x d x

(t - 1)^{2}

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 5, y^{^{'}} (0) = - 1

\frac{\partial}{\partial x} (x cos (x))

\int - 4 x^{6} e^{x} d x

What is the integral of sin(18x)cos(13x) ?
The integral of sin(18x)cos(13x) is 1/2 (-1/31 cos(31x)-1/5 cos(5x))+C