What is the integral of sin(24x)cos(15x) ?

The integral of sin(24x)cos(15x) is 1/2 (-1/39 cos(39x)-1/9 cos(9x))+C

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integral of sin(24x)cos(15x)

\int sin (24 x) cos (15 x) d x

\frac{1}{2} (- \frac{1}{39} cos (39 x) - \frac{1}{9} cos (9 x)) + C

Solution steps

\int sin (24 x) cos (15 x) d x

Use the following identity:

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

= \int \frac{sin ( 24 x + 15 x ) + sin ( 24 x - 15 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 (24 x + 15 x) + sin (24 x - 15 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (39 x) + sin (9 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 (39 x) d x + \int sin (9 x) d x)

\int sin (39 x) d x = - \frac{1}{39} cos (39 x)

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

= \frac{1}{2} (- \frac{1}{39} cos (39 x) - \frac{1}{9} cos (9 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{39} cos (39 x) - \frac{1}{9} cos (9 x)) + C

\frac{d}{d x} (\frac{8}{x - 2})

\int x^{2} \cdot e^{x^{3} + 4} d x

\frac{d}{d x} (r (θ, x) cos (θ))

y = 4 arcsin (5) x^{5}

(1 + x)^{- 1}

What is the integral of sin(24x)cos(15x) ?
The integral of sin(24x)cos(15x) is 1/2 (-1/39 cos(39x)-1/9 cos(9x))+C