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

The integral from 0 to 2*pi of sin(x) is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{0}^{2 \cdot π} sin (x) d x

0

Solution steps

\int_{0}^{2 π} sin (x) d x

Use the common integral:

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

= [- cos (x)]_{0}^{2 π}

Compute the boundaries:

0

= 0

\int_{0}^{2 π} cos^{2} (v) d v

\int_{0}^{2} (x^{2} + 4 x) d x

\int_{0}^{5} 5 x^{2} d x

\int_{0}^{3} 8 e^{2 x} d x

\int_{- π}^{0} x \cdot cos (n x) d x

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