What is the integral of sin((-5x)/2)cos(4x) ?

The integral of sin((-5x)/2)cos(4x) is -1/39 (-3cos((13x)/2)+13cos((3x)/2))+C

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integral of sin((-5x)/2)cos(4x)

\int sin (\frac{- 5 x}{2}) cos (4 x) d x

- \frac{1}{39} (- 3 cos (\frac{13 x}{2}) + 13 cos (\frac{3 x}{2})) + C

Solution steps

\int sin (\frac{- 5 x}{2}) cos (4 x) d x

Apply u-substitution

: \int - \frac{1}{4} sin (\frac{5 u}{8}) cos (u) d u

= \int - \frac{1}{4} sin (\frac{5 u}{8}) cos (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int sin (\frac{5 u}{8}) cos (u) d u

Use the following identity:

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

= - \frac{1}{4} \cdot \int \frac{sin ( \frac{5 u}{8} + u ) + sin ( \frac{5 u}{8} - u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \frac{1}{2} \cdot \int sin (\frac{5 u}{8} + u) + sin (\frac{5 u}{8} - u) d u

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}{4} \cdot \frac{1}{2} (\int sin (\frac{5 u}{8} + u) d u + \int sin (\frac{5 u}{8} - u) d u)

\int sin (\frac{5 u}{8} + u) d u = - \frac{8}{13} cos (\frac{13 u}{8})

\int sin (\frac{5 u}{8} - u) d u = \frac{8}{3} cos (\frac{3 u}{8})

= - \frac{1}{4} \cdot \frac{1}{2} (- \frac{8}{13} cos (\frac{13 u}{8}) + \frac{8}{3} cos (\frac{3 u}{8}))

Substitute back

u = 4 x

= - \frac{1}{4} \cdot \frac{1}{2} (- \frac{8}{13} cos (\frac{13 \cdot 4 x}{8}) + \frac{8}{3} cos (\frac{3 \cdot 4 x}{8}))

Simplify

- \frac{1}{4} \cdot \frac{1}{2} (- \frac{8}{13} cos (\frac{13 \cdot 4 x}{8}) + \frac{8}{3} cos (\frac{3 \cdot 4 x}{8})) : - \frac{1}{39} (- 3 cos (\frac{13 x}{2}) + 13 cos (\frac{3 x}{2}))

= - \frac{1}{39} (- 3 cos (\frac{13 x}{2}) + 13 cos (\frac{3 x}{2}))

Add a constant to the solution

= - \frac{1}{39} (- 3 cos (\frac{13 x}{2}) + 13 cos (\frac{3 x}{2})) + C

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What is the integral of sin((-5x)/2)cos(4x) ?
The integral of sin((-5x)/2)cos(4x) is -1/39 (-3cos((13x)/2)+13cos((3x)/2))+C