What is the integral of sin(-10x)cos(9x) ?

The integral of sin(-10x)cos(9x) is 1/2 (cos(x)+1/19 cos(19x))+C

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integral of sin(-10x)cos(9x)

\int sin (- 10 x) cos (9 x) d x

\frac{1}{2} (cos (x) + \frac{1}{19} cos (19 x)) + C

Solution steps

\int sin (- 10 x) cos (9 x) d x

Use the following identity:

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

= \int \frac{sin ( - 10 x + 9 x ) + sin ( - 10 x - 9 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 (- 10 x + 9 x) + sin (- 10 x - 9 x) d x

Simplify

sin (- 10 x + 9 x) + sin (- 10 x - 9 x) : - sin (x) - sin (19 x)

= \frac{1}{2} \cdot \int - sin (x) - sin (19 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 (x) d x - \int sin (19 x) d x)

\int sin (x) d x = - cos (x)

\int sin (19 x) d x = - \frac{1}{19} cos (19 x)

= \frac{1}{2} (- (- cos (x)) - (- \frac{1}{19} cos (19 x)))

Simplify

= \frac{1}{2} (cos (x) + \frac{1}{19} cos (19 x))

Add a constant to the solution

= \frac{1}{2} (cos (x) + \frac{1}{19} cos (19 x)) + C

What is the integral of sin(-10x)cos(9x) ?
The integral of sin(-10x)cos(9x) is 1/2 (cos(x)+1/19 cos(19x))+C