What is the integral of sin(16x)cos(7x) ?

The integral of sin(16x)cos(7x) is 1/2 (-1/23 cos(23x)-1/9 cos(9x))+C

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integral of sin(16x)cos(7x)

\int sin (16 x) cos (7 x) d x

\frac{1}{2} (- \frac{1}{23} cos (23 x) - \frac{1}{9} cos (9 x)) + C

Solution steps

\int sin (16 x) cos (7 x) d x

Use the following identity:

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

= \int \frac{sin ( 16 x + 7 x ) + sin ( 16 x - 7 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 (16 x + 7 x) + sin (16 x - 7 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 (16 x + 7 x) d x + \int sin (16 x - 7 x) d x)

\int sin (16 x + 7 x) d x = - \frac{1}{23} cos (23 x)

\int sin (16 x - 7 x) d x = - \frac{1}{9} cos (9 x)

= \frac{1}{2} (- \frac{1}{23} cos (23 x) - \frac{1}{9} cos (9 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{23} cos (23 x) - \frac{1}{9} cos (9 x)) + C

\frac{\partial}{\partial x} (x^{7} y^{6})

\int 5 x - 1 d x

\frac{d y}{d x} = 8 - \frac{y}{50}, y (0) = 50

y^{^{''}} = 6 t

f (x) = \frac{1}{x ^{8}}

What is the integral of sin(16x)cos(7x) ?
The integral of sin(16x)cos(7x) is 1/2 (-1/23 cos(23x)-1/9 cos(9x))+C