What is the integral of 16x^3cos(2-x^4) ?

The integral of 16x^3cos(2-x^4) is -4sin(2-x^4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 16x^3cos(2-x^4)

\int 16 x^{3} cos (2 - x^{4}) d x

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

Solution steps

\int 16 x^{3} cos (2 - x^{4}) d x

Take the constant out:

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

= 16 \cdot \int x^{3} cos (2 - x^{4}) d x

Apply u-substitution

: \int - \frac{cos ( u )}{4} d u

= 16 \cdot \int - \frac{cos ( u )}{4} d u

Take the constant out:

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

= 16 (- \frac{1}{4} \cdot \int cos (u) d u)

Use the common integral:

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

= 16 (- \frac{1}{4} sin (u))

Substitute back

u = 2 - x^{4}

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

Simplify

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

= - 4 sin (2 - x^{4})

Add a constant to the solution

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

\frac{d}{d u} (cos (u))

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = - 3

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

x \to - 1 lim (8 x^{4} - 3 x^{3} + 7 x)

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

What is the integral of 16x^3cos(2-x^4) ?
The integral of 16x^3cos(2-x^4) is -4sin(2-x^4)+C