What is the integral of sin(4y)cos(5y) ?

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

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integral of sin(4y)cos(5y)

\int sin (4 y) cos (5 y) d y

\frac{1}{2} (- \frac{1}{9} cos (9 y) + cos (y)) + C

Solution steps

\int sin (4 y) cos (5 y) d y

Use the following identity:

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

= \int \frac{sin ( 4 y + 5 y ) + sin ( 4 y - 5 y )}{2} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (4 y + 5 y) + sin (4 y - 5 y) d y

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 (4 y + 5 y) d y + \int sin (4 y - 5 y) d y)

\int sin (4 y + 5 y) d y = - \frac{1}{9} cos (9 y)

\int sin (4 y - 5 y) d y = cos (y)

= \frac{1}{2} (- \frac{1}{9} cos (9 y) + cos (y))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{9} cos (9 y) + cos (y)) + C

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