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

The integral from 0 to 2pi of sin(nx) 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 2pi of sin(nx)

\int_{0}^{2 π} sin (n x) d x

0

Solution steps

\int_{0}^{2 π} sin (n x) d x

Apply u-substitution

: \int_{0}^{n \cdot 2 π} \frac{sin ( u )}{n} d u

= \int_{0}^{n \cdot 2 π} \frac{sin ( u )}{n} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{n} [- cos (u)]_{0}^{n \cdot 2 π}

Compute the boundaries:

0

= \frac{1}{n} \cdot 0

Simplify

= 0

\int_{0}^{2} 3 x^{2} - 2 x + 7 e^{x} d x

\int_{- 2}^{1} \frac{1}{x} d x

\int_{1}^{5} 5 d x

\int_{- 2}^{3} x^{2} - 1 d x

\int_{0}^{\frac{π}{3}} 3 cos (x) + 1 d x

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