What is the integral of cos(14x)sin(8x) ?

The integral of cos(14x)sin(8x) is 1/2 (-1/22 cos(22x)+1/6 cos(6x))+C

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integral of cos(14x)sin(8x)

\int cos (14 x) sin (8 x) d x

\frac{1}{2} (- \frac{1}{22} cos (22 x) + \frac{1}{6} cos (6 x)) + C

Solution steps

\int cos (14 x) sin (8 x) d x

Use the following identity:

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

= \int \frac{sin ( 8 x + 14 x ) + sin ( 8 x - 14 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 (8 x + 14 x) + sin (8 x - 14 x) d x

Simplify

sin (8 x + 14 x) + sin (8 x - 14 x) : sin (22 x) - sin (6 x)

= \frac{1}{2} \cdot \int sin (22 x) - sin (6 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 (22 x) d x - \int sin (6 x) d x)

\int sin (22 x) d x = - \frac{1}{22} cos (22 x)

\int sin (6 x) d x = - \frac{1}{6} cos (6 x)

= \frac{1}{2} (- \frac{1}{22} cos (22 x) - (- \frac{1}{6} cos (6 x)))

Simplify

= \frac{1}{2} (- \frac{1}{22} cos (22 x) + \frac{1}{6} cos (6 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{22} cos (22 x) + \frac{1}{6} cos (6 x)) + C

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What is the integral of cos(14x)sin(8x) ?
The integral of cos(14x)sin(8x) is 1/2 (-1/22 cos(22x)+1/6 cos(6x))+C