What is the integral of sin(-7x)cos(6x) ?

The integral of sin(-7x)cos(6x) is 1/2 (1/13 cos(13x)+cos(x))+C

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integral of sin(-7x)cos(6x)

\int sin (- 7 x) cos (6 x) d x

\frac{1}{2} (\frac{1}{13} cos (13 x) + cos (x)) + C

Solution steps

\int sin (- 7 x) cos (6 x) d x

Use the following identity:

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

= \int \frac{sin ( - 7 x + 6 x ) + sin ( - 7 x - 6 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 (- 7 x + 6 x) + sin (- 7 x - 6 x) d x

Simplify

sin (- 7 x + 6 x) + sin (- 7 x - 6 x) : - sin (13 x) - sin (x)

= \frac{1}{2} \cdot \int - sin (13 x) - sin (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 (13 x) d x - \int sin (x) d x)

\int sin (13 x) d x = - \frac{1}{13} cos (13 x)

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

= \frac{1}{2} (- (- \frac{1}{13} cos (13 x)) - (- cos (x)))

Simplify

= \frac{1}{2} (\frac{1}{13} cos (13 x) + cos (x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{13} cos (13 x) + cos (x)) + C

- 2744 - x^{3}

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What is the integral of sin(-7x)cos(6x) ?
The integral of sin(-7x)cos(6x) is 1/2 (1/13 cos(13x)+cos(x))+C