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

The integral from 0 to 2pi of xcos(nx) is (2pisin(2pin))/n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 2pi of xcos(nx)

\int_{0}^{2 π} x cos (n x) d x

\frac{2 π sin ( 2 πn )}{n}

Solution steps

\int_{0}^{2 π} x cos (n x) d x

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

= \frac{1}{n ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{2 πn}

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

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

Simplify

= \frac{1}{n ^{2}} [u sin (u) + cos (u)]_{0}^{2 πn}

Compute the boundaries:

2 πn sin (2 πn)

= \frac{1}{n ^{2}} \cdot 2 πn sin (2 πn)

Simplify

= \frac{2 π sin ( 2 πn )}{n}

What is the integral from 0 to 2pi of xcos(nx) ?
The integral from 0 to 2pi of xcos(nx) is (2pisin(2pin))/n