What is the integral from 0 to 2x of sin(2x) ?

The integral from 0 to 2x of sin(2x) is 1/2 (-cos(4x)+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2x of sin(2x)

\int_{0}^{2 x} sin (2 x) d x

\frac{1}{2} (- cos (4 x) + 1)

Solution steps

\int_{0}^{2 x} sin (2 x) d x

Apply u-substitution

: \int_{0}^{4 x} sin (u) \frac{1}{2} d u

= \int_{0}^{4 x} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{4 x} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{4 x}

Compute the boundaries:

- cos (4 x) + 1

= \frac{1}{2} (- cos (4 x) + 1)

(6 x^{2} + 8)^{6}

\int (e^{x} sin (y)) d x

\int \frac{e ^{x}}{( 1 - e ^{x} ) ^{5}} d x

\int \frac{x + 2}{x - 2} d x

f (x) = \frac{1}{x + 8}, at x = 5

What is the integral from 0 to 2x of sin(2x) ?
The integral from 0 to 2x of sin(2x) is 1/2 (-cos(4x)+1)