What is the integral of cos(3x-2)sin(4-5x) ?

The integral of cos(3x-2)sin(4-5x) is 1/2 (1/2 cos(-2x+2)+1/8 cos(-8x+6))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cos(3x-2)sin(4-5x)

\int cos (3 x - 2) sin (4 - 5 x) d x

\frac{1}{2} (\frac{1}{2} cos (- 2 x + 2) + \frac{1}{8} cos (- 8 x + 6)) + C

Solution steps

\int cos (3 x - 2) sin (4 - 5 x) d x

Use the following identity:

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

= \int \frac{sin ( 4 - 5 x + 3 x - 2 ) + sin ( 4 - 5 x - ( 3 x - 2 ) )}{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 - 5 x + 3 x - 2) + sin (4 - 5 x - (3 x - 2)) 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 - 5 x + 3 x - 2) d x + \int sin (4 - 5 x - (3 x - 2)) d x)

\int sin (4 - 5 x + 3 x - 2) d x = \frac{1}{2} cos (- 2 x + 2)

\int sin (4 - 5 x - (3 x - 2)) d x = \frac{1}{8} cos (- 8 x + 6)

= \frac{1}{2} (\frac{1}{2} cos (- 2 x + 2) + \frac{1}{8} cos (- 8 x + 6))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} cos (- 2 x + 2) + \frac{1}{8} cos (- 8 x + 6)) + C

sin (x), [\frac{11}{9} π, \frac{12}{8} π]

\int tan^{3} (x) + tan (x) d x

(1.1) (2.4)

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

\frac{1}{( 7 + x ) ^{3}}

What is the integral of cos(3x-2)sin(4-5x) ?
The integral of cos(3x-2)sin(4-5x) is 1/2 (1/2 cos(-2x+2)+1/8 cos(-8x+6))+C