What is the integral of sin(8x)cos(5x) ?

The integral of sin(8x)cos(5x) is 1/2 (-1/13 cos(13x)-1/3 cos(3x))+C

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integral of sin(8x)cos(5x)

\int sin (8 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (8 x) cos (5 x) d x

Use the following identity:

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

= \int \frac{sin ( 8 x + 5 x ) + sin ( 8 x - 5 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 (8 x + 5 x) + sin (8 x - 5 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (13 x) + sin (3 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 (3 x) d x)

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

\int sin (3 x) d x = - \frac{1}{3} cos (3 x)

= \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x)) + C

\frac{d}{d x} (\frac{7 - 2 x}{x ^{2} + 8})

y = 1.25 x, x = 8.5 - y^{2}

\frac{d}{d x} (48 x (4 x^{2} + 9)^{5})

f (x) = \frac{4 x - 3}{3 x + 2} + x^{3}

\int \frac{x ^{2} + 4 x - 5}{x ^{3} - 1} d x

What is the integral of sin(8x)cos(5x) ?
The integral of sin(8x)cos(5x) is 1/2 (-1/13 cos(13x)-1/3 cos(3x))+C