What is the integral of sin(22x)cos(15x) ?

The integral of sin(22x)cos(15x) is 1/2 (-1/37 cos(37x)-1/7 cos(7x))+C

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integral of sin(22x)cos(15x)

\int sin (22 x) cos (15 x) d x

\frac{1}{2} (- \frac{1}{37} cos (37 x) - \frac{1}{7} cos (7 x)) + C

Solution steps

\int sin (22 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 ( 22 x + 15 x ) + sin ( 22 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 (22 x + 15 x) + sin (22 x - 15 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (37 x) + sin (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 (37 x) d x + \int sin (7 x) d x)

\int sin (37 x) d x = - \frac{1}{37} cos (37 x)

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

= \frac{1}{2} (- \frac{1}{37} cos (37 x) - \frac{1}{7} cos (7 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{37} cos (37 x) - \frac{1}{7} cos (7 x)) + C

y = 2 x - x^{2}, y = - x

y = sin (h) 3 x

y = (3 x^{2} + 2) (2 x - 7)

\int_{1}^{e^{4}} \frac{1}{x ( 1 + ln ( x ) )} d x

\int y^{2} \cdot cos (y) d y

What is the integral of sin(22x)cos(15x) ?
The integral of sin(22x)cos(15x) is 1/2 (-1/37 cos(37x)-1/7 cos(7x))+C