What is the integral from 1 to pi+1 of sin(2x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to pi+1 of sin(2x)

\int_{1}^{π + 1} sin (2 x) d x

\frac{1}{2} (- cos (2 (π + 1)) + cos (2))

Decimal

0

Solution steps

\int_{1}^{π + 1} sin (2 x) d x

Apply u-substitution

= \int_{2}^{2 (π + 1)} 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_{2}^{2 (π + 1)} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{2}^{2 (π + 1)}

Compute the boundaries:

- cos (2 (π + 1)) + cos (2)

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

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