What is the integral of sin(11z)cos(6z) ?

The integral of sin(11z)cos(6z) is 1/2 (-1/17 cos(17z)-1/5 cos(5z))+C

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integral of sin(11z)cos(6z)

\int sin (11 z) cos (6 z) d z

\frac{1}{2} (- \frac{1}{17} cos (17 z) - \frac{1}{5} cos (5 z)) + C

Solution steps

\int sin (11 z) cos (6 z) d z

Use the following identity:

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

= \int \frac{sin ( 11 z + 6 z ) + sin ( 11 z - 6 z )}{2} d z

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sin (11 z + 6 z) + sin (11 z - 6 z) d z

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 (11 z + 6 z) d z + \int sin (11 z - 6 z) d z)

\int sin (11 z + 6 z) d z = - \frac{1}{17} cos (17 z)

\int sin (11 z - 6 z) d z = - \frac{1}{5} cos (5 z)

= \frac{1}{2} (- \frac{1}{17} cos (17 z) - \frac{1}{5} cos (5 z))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{17} cos (17 z) - \frac{1}{5} cos (5 z)) + C

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What is the integral of sin(11z)cos(6z) ?
The integral of sin(11z)cos(6z) is 1/2 (-1/17 cos(17z)-1/5 cos(5z))+C