What is the integral of sin(22x)cos(17x) ?

The integral of sin(22x)cos(17x) is 1/2 (-1/39 cos(39x)-1/5 cos(5x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(22x)cos(17x)

\int sin (22 x) cos (17 x) d x

\frac{1}{2} (- \frac{1}{39} cos (39 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

\int sin (22 x) cos (17 x) d x

Use the following identity:

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

= \int \frac{sin ( 22 x + 17 x ) + sin ( 22 x - 17 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 + 17 x) + sin (22 x - 17 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (39 x) + sin (5 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 (39 x) d x + \int sin (5 x) d x)

\int sin (39 x) d x = - \frac{1}{39} cos (39 x)

\int sin (5 x) d x = - \frac{1}{5} cos (5 x)

= \frac{1}{2} (- \frac{1}{39} cos (39 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{39} cos (39 x) - \frac{1}{5} cos (5 x)) + C

\int (x + 2) 5 - x d x

\frac{d}{d x} ((x) \cdot sin (x))

\int x e^{x^{2}} sin (x^{2}) d x

y = lo g_{3} (x^{2} + 1)

\frac{d y}{d x} = \frac{26}{x}

What is the integral of sin(22x)cos(17x) ?
The integral of sin(22x)cos(17x) is 1/2 (-1/39 cos(39x)-1/5 cos(5x))+C