What is the integral from 0 to pi of x/pi cos(nx) ?

The integral from 0 to pi of x/pi cos(nx) is ((-1)^n-1)/(pin^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi of x/pi cos(nx)

\int_{0}^{π} \frac{x}{π} cos (n x) d x

\frac{( - 1 ) ^{n} - 1}{π n ^{2}}

Solution steps

\int_{0}^{π} \frac{x}{π} cos (n x) d x

Take the constant out:

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

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

\frac{1}{π} \cdot \frac{1}{n ^{2}} [u sin (u) - (- cos (u))]_{0}^{πn} : \frac{1}{π n ^{2}} [u sin (u) + cos (u)]_{0}^{πn}

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

Compute the boundaries:

(- 1)^{n} - 1

= \frac{1}{π n ^{2}} ((- 1)^{n} - 1)

Simplify

= \frac{( - 1 ) ^{n} - 1}{π n ^{2}}

\int_{1}^{2} (x^{2} + 4) d x

\int_{0}^{2 π} 6 x 2 d x

\int_{1}^{2} (\frac{x ^{2}}{4} + 1) d x

\int_{0}^{10} x e^{- x^{2}} d x

\int_{3}^{5} \frac{1}{∣ x - 1 ∣} d x

What is the integral from 0 to pi of x/pi cos(nx) ?
The integral from 0 to pi of x/pi cos(nx) is ((-1)^n-1)/(pin^2)