What is the integral of 8x^3-cos(4x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 8x^3-cos(4x)

\int 8 x^{3} - cos (4 x) d x

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

Solution steps

\int 8 x^{3} - 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 8 x^{3} d x - \int cos (4 x) d x

\int 8 x^{3} d x = 2 x^{4}

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

= 2 x^{4} - \frac{1}{4} sin (4 x)

Add a constant to the solution

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

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