What is the integral of sin(8x)cos(12x) ?

The integral of sin(8x)cos(12x) is 1/2 (-1/20 cos(20x)+1/4 cos(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sin(8x)cos(12x)

\int sin (8 x) cos (12 x) d x

\frac{1}{2} (- \frac{1}{20} cos (20 x) + \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (8 x) cos (12 x) d x

Use the following identity:

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

= \int \frac{sin ( 8 x + 12 x ) + sin ( 8 x - 12 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 (8 x + 12 x) + sin (8 x - 12 x) d x

Simplify

sin (8 x + 12 x) + sin (8 x - 12 x) : sin (20 x) - sin (4 x)

= \frac{1}{2} \cdot \int sin (20 x) - sin (4 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 (20 x) d x - \int sin (4 x) d x)

\int sin (20 x) d x = - \frac{1}{20} cos (20 x)

\int sin (4 x) d x = - \frac{1}{4} cos (4 x)

= \frac{1}{2} (- \frac{1}{20} cos (20 x) - (- \frac{1}{4} cos (4 x)))

Simplify

= \frac{1}{2} (- \frac{1}{20} cos (20 x) + \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{20} cos (20 x) + \frac{1}{4} cos (4 x)) + C

What is the integral of sin(8x)cos(12x) ?
The integral of sin(8x)cos(12x) is 1/2 (-1/20 cos(20x)+1/4 cos(4x))+C