What is the integral of (4x+8)sin(x^2+4x-2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x+8)sin(x^2+4x-2)

\int (4 x + 8) sin (x^{2} + 4 x - 2) d x

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

Solution steps

\int (4 x + 8) sin (x^{2} + 4 x - 2) d x

Apply u-substitution

: \int 2 sin (u) d u

= \int 2 sin (u) d u

Take the constant out:

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

= 2 \cdot \int sin (u) d u

Use the common integral:

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

= 2 (- cos (u))

Substitute back

u = x^{2} + 4 x - 2

= 2 (- cos (x^{2} + 4 x - 2))

Simplify

= - 2 cos (x^{2} + 4 x - 2)

Add a constant to the solution

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

y = ln (x + x^{2} - 3)

\frac{d y}{d x} + 5 y - 20 = 0

n = 1 \sum \infty 7

x^{2}, - x + 2

\int \frac{1}{1 + 3 x} d x

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