What is the integral of sin(5x)*cos(2x) ?

The integral of sin(5x)*cos(2x) is 1/2 (-1/7 cos(7x)-1/3 cos(3x))+C

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integral of sin(5x)*cos(2x)

\int sin (5 x) \cdot cos (2 x) d x

\frac{1}{2} (- \frac{1}{7} cos (7 x) - \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (5 x) cos (2 x) d x

Use the following identity:

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

= \int \frac{sin ( 5 x + 2 x ) + sin ( 5 x - 2 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 + 2 x) + sin (5 x - 2 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 + 2 x) d x + \int sin (5 x - 2 x) d x)

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

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

= \frac{1}{2} (- \frac{1}{7} cos (7 x) - \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{7} cos (7 x) - \frac{1}{3} cos (3 x)) + C

f (t) = ln (t^{9} + 5)

y = 4 x - x^{2}, (1, 3)

\frac{d y}{d x} + 8 y = 10

f (x) = 3 x - 4 x^{2}, (- 3, - 45)

\frac{d}{d x} ((x - 3)^{3} (2 x - 1)^{2})

What is the integral of sin(5x)*cos(2x) ?
The integral of sin(5x)*cos(2x) is 1/2 (-1/7 cos(7x)-1/3 cos(3x))+C