What is the integral of sin(9x)cos(2x) ?

The integral of sin(9x)cos(2x) is 1/2 (-1/11 cos(11x)-1/7 cos(7x))+C

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integral of sin(9x)cos(2x)

\int sin (9 x) cos (2 x) d x

\frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{7} cos (7 x)) + C

Solution steps

\int sin (9 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 ( 9 x + 2 x ) + sin ( 9 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 (9 x + 2 x) + sin (9 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 (9 x + 2 x) d x + \int sin (9 x - 2 x) d x)

\int sin (9 x + 2 x) d x = - \frac{1}{11} cos (11 x)

\int sin (9 x - 2 x) d x = - \frac{1}{7} cos (7 x)

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{7} cos (7 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{7} cos (7 x)) + C

\int 15 cot^{4} (x) d x

\frac{\partial}{\partial x} (- 2 y cos^{2} (x))

\frac{d y}{d x} + y = x^{2} + 1

\frac{\partial}{\partial x} (y^{2} - cos (x y))

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

What is the integral of sin(9x)cos(2x) ?
The integral of sin(9x)cos(2x) is 1/2 (-1/11 cos(11x)-1/7 cos(7x))+C