What is the integral of sin(3x)cos(5x) ?

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

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integral of sin(3x)cos(5x)

\int sin (3 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{2} cos (2 x)) + C

Solution steps

\int sin (3 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 ( 3 x + 5 x ) + sin ( 3 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 (3 x + 5 x) + sin (3 x - 5 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 (3 x + 5 x) d x + \int sin (3 x - 5 x) d x)

\int sin (3 x + 5 x) d x = - \frac{1}{8} cos (8 x)

\int sin (3 x - 5 x) d x = \frac{1}{2} cos (2 x)

= \frac{1}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{8} cos (8 x) + \frac{1}{2} cos (2 x)) + C

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\int_{4}^{6} \frac{1}{x ^{2} - 1} d x

\frac{\partial}{\partial x} (t^{8} e^{- x})

\frac{d}{d x} (\frac{x + 1}{x - 1})

\frac{\partial}{\partial x} (tan (5 + 2 x^{2} y^{3} z^{2}))

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