What is the integral of sin(6x)cos(4x) ?

The integral of sin(6x)cos(4x) is 1/2 (-1/10 cos(10x)-1/2 cos(2x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sin(6x)cos(4x)

\int sin (6 x) cos (4 x) d x

\frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{2} cos (2 x)) + C

Solution steps

\int sin (6 x) cos (4 x) d x

Use the following identity:

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

= \int \frac{sin ( 6 x + 4 x ) + sin ( 6 x - 4 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 (6 x + 4 x) + sin (6 x - 4 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (10 x) + sin (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 (10 x) d x + \int sin (2 x) d x)

\int sin (10 x) d x = - \frac{1}{10} cos (10 x)

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

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{2} cos (2 x)) + C

What is the integral of sin(6x)cos(4x) ?
The integral of sin(6x)cos(4x) is 1/2 (-1/10 cos(10x)-1/2 cos(2x))+C