What is the integral of sin(9x)cos(6x) ?

The integral of sin(9x)cos(6x) is 1/2 (-1/15 cos(15x)-1/3 cos(3x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(9x)cos(6x)

\int sin (9 x) cos (6 x) d x

\frac{1}{2} (- \frac{1}{15} cos (15 x) - \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (9 x) cos (6 x) d x

Use the following identity:

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

= \int \frac{sin ( 9 x + 6 x ) + sin ( 9 x - 6 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 + 6 x) + sin (9 x - 6 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (15 x) + sin (3 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 (15 x) d x + \int sin (3 x) d x)

\int sin (15 x) d x = - \frac{1}{15} cos (15 x)

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

= \frac{1}{2} (- \frac{1}{15} cos (15 x) - \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{15} cos (15 x) - \frac{1}{3} cos (3 x)) + C

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

\frac{d}{d θ} (cos^{2} (θ))

y^{3} - (2 x + 8) + 3 x y^{2} y^{^{'}} = 0

\frac{\partial}{\partial y} (3 x^{2} + 4 x y + y^{2})

\frac{\partial}{\partial x} (6 x y)

What is the integral of sin(9x)cos(6x) ?
The integral of sin(9x)cos(6x) is 1/2 (-1/15 cos(15x)-1/3 cos(3x))+C