What is the integral of sin(18x)cos(11x) ?

The integral of sin(18x)cos(11x) is 1/2 (-1/29 cos(29x)-1/7 cos(7x))+C

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integral of sin(18x)cos(11x)

\int sin (18 x) cos (11 x) d x

\frac{1}{2} (- \frac{1}{29} cos (29 x) - \frac{1}{7} cos (7 x)) + C

Solution steps

\int sin (18 x) cos (11 x) d x

Use the following identity:

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

= \int \frac{sin ( 18 x + 11 x ) + sin ( 18 x - 11 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 + 11 x) + sin (18 x - 11 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (29 x) + sin (7 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 (29 x) d x + \int sin (7 x) d x)

\int sin (29 x) d x = - \frac{1}{29} cos (29 x)

\int sin (7 x) d x = - \frac{1}{7} cos (7 x)

= \frac{1}{2} (- \frac{1}{29} cos (29 x) - \frac{1}{7} cos (7 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{29} cos (29 x) - \frac{1}{7} cos (7 x)) + C

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What is the integral of sin(18x)cos(11x) ?
The integral of sin(18x)cos(11x) is 1/2 (-1/29 cos(29x)-1/7 cos(7x))+C