What is the integral of sin(5x)cos(7x) ?

The integral of sin(5x)cos(7x) is 1/2 (-1/12 cos(12x)+1/2 cos(2x))+C

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integral of sin(5x)cos(7x)

\int sin (5 x) cos (7 x) d x

\frac{1}{2} (- \frac{1}{12} cos (12 x) + \frac{1}{2} cos (2 x)) + C

Solution steps

\int sin (5 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 ( 5 x + 7 x ) + sin ( 5 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 (5 x + 7 x) + sin (5 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 (5 x + 7 x) d x + \int sin (5 x - 7 x) d x)

\int sin (5 x + 7 x) d x = - \frac{1}{12} cos (12 x)

\int sin (5 x - 7 x) d x = \frac{1}{2} cos (2 x)

= \frac{1}{2} (- \frac{1}{12} cos (12 x) + \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{12} cos (12 x) + \frac{1}{2} cos (2 x)) + C

\frac{5}{x}, \frac{5}{x ^{2}}, 1, 4

y = - 3 x + 2, x = 2

\int (\frac{x + 9}{x}) d x

y^{^{''}} + 2 y^{^{'}} + y = 3 e^{- x} x + 1

f (t) = ln (t^{2} + 1)

What is the integral of sin(5x)cos(7x) ?
The integral of sin(5x)cos(7x) is 1/2 (-1/12 cos(12x)+1/2 cos(2x))+C