What is the integral of-8x^3cos(5x^4+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-8x^3cos(5x^4+1)

\int - 8 x^{3} cos (5 x^{4} + 1) d x

- \frac{2}{5} sin (5 x^{4} + 1) + C

Solution steps

\int - 8 x^{3} cos (5 x^{4} + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 8 \cdot \int x^{3} cos (5 x^{4} + 1) d x

Apply u-substitution

: \int \frac{cos ( u )}{20} d u

= - 8 \cdot \int \frac{cos ( u )}{20} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 8 \cdot \frac{1}{20} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= - 8 \cdot \frac{1}{20} sin (u)

Substitute back

u = 5 x^{4} + 1

= - 8 \cdot \frac{1}{20} sin (5 x^{4} + 1)

Simplify

- 8 \cdot \frac{1}{20} sin (5 x^{4} + 1) : - \frac{2}{5} sin (5 x^{4} + 1)

= - \frac{2}{5} sin (5 x^{4} + 1)

Add a constant to the solution

= - \frac{2}{5} sin (5 x^{4} + 1) + C

3 e^{6 x}

\frac{\partial}{\partial x} (16 x - 4 y - 6 y x)

\int_{- 1}^{2} (2 - 5 x + \frac{1}{2} x^{2}) d x

f (x) = (x + 9)^{\frac{1}{2}}

f (x) = - x^{2} - x

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