What is the integral of (2x^2-x)cos(4x) ?

The integral of (2x^2-x)cos(4x) is 1/16 (8x^2sin(4x)-sin(4x)+4xcos(4x))-1/16 (4xsin(4x)+cos(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^2-x)cos(4x)

\int (2 x^{2} - x) cos (4 x) d x

\frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) - \frac{1}{16} (4 x sin (4 x) + cos (4 x)) + C

Solution steps

\int (2 x^{2} - x) cos (4 x) d x

Expand

(2 x^{2} - x) cos (4 x) : 2 x^{2} cos (4 x) - x cos (4 x)

= \int 2 x^{2} cos (4 x) - x cos (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

= \int 2 x^{2} cos (4 x) d x - \int x cos (4 x) d x

\int 2 x^{2} cos (4 x) d x = \frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x))

\int x cos (4 x) d x = \frac{1}{16} (4 x sin (4 x) + cos (4 x))

= \frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) - \frac{1}{16} (4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{1}{16} (8 x^{2} sin (4 x) - sin (4 x) + 4 x cos (4 x)) - \frac{1}{16} (4 x sin (4 x) + cos (4 x)) + C

\int (1 - 2 x^{2}) e^{- x^{2}} d x

\int (6 x + 3) e^{- 2 x} d x

\int e^{c o s (3 x)} sin (3 x) d x

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

\int y^{2} - 49 d y

What is the integral of (2x^2-x)cos(4x) ?
The integral of (2x^2-x)cos(4x) is 1/16 (8x^2sin(4x)-sin(4x)+4xcos(4x))-1/16 (4xsin(4x)+cos(4x))+C