What is the integral from 0 to 2pi of θcos(nθ) ?

The integral from 0 to 2pi of θcos(nθ) is (2pisin(2pin))/n

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 θcos(nθ)

\int_{0}^{2 π} θ cos (n θ) d θ

\frac{2 π sin ( 2 πn )}{n}

Solution steps

\int_{0}^{2 π} θ cos (n θ) d θ

Apply u-substitution

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

= \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

: [u sin (u) - \int sin (u) d u]_{0}^{2 πn}

= \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}

\int_{0}^{π} (3 - sin (x))^{6} d x

\int_{0}^{3} y 9 - y^{2} d y

\int_{4}^{1} x d x

\int_{6}^{7} 13 x - x^{2} - 42 d x

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

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