What is the integral of sin(4x)cos(5x) ?

The integral of sin(4x)cos(5x) is 1/2 (-1/9 cos(9x)+cos(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(4x)cos(5x)

\int sin (4 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{9} cos (9 x) + cos (x)) + C

Solution steps

\int sin (4 x) cos (5 x) d x

Use the following identity:

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

= \int \frac{sin ( 4 x + 5 x ) + sin ( 4 x - 5 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 (4 x + 5 x) + sin (4 x - 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 (4 x + 5 x) d x + \int sin (4 x - 5 x) d x)

\int sin (4 x + 5 x) d x = - \frac{1}{9} cos (9 x)

\int sin (4 x - 5 x) d x = cos (x)

= \frac{1}{2} (- \frac{1}{9} cos (9 x) + cos (x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{9} cos (9 x) + cos (x)) + C

\int_{1}^{2} (2 - x)^{\frac{- 2}{3}} d x

\int \frac{1}{1 - 2 y} d y

\frac{d}{d v} (\frac{v ^{3} - 2 v v}{v})

7 e^{y}

\int \frac{( x ^{2} )}{6 x ^{2} - 49} d x

What is the integral of sin(4x)cos(5x) ?
The integral of sin(4x)cos(5x) is 1/2 (-1/9 cos(9x)+cos(x))+C