What is the integral from 0 to pi of (pi-t)sin(nt) ?

The integral from 0 to pi of (pi-t)sin(nt) is pi/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 pi of (pi-t)sin(nt)

\int_{0}^{π} (π - t) sin (n t) d t

\frac{π}{n}

Solution steps

\int_{0}^{π} (π - t) sin (n t) d t

Apply Integration By Parts

: [\frac{1}{n} (- cos (n t) (π - t) - n \cdot \int \frac{1}{n} cos (n t) d t)]_{0}^{π}

= [\frac{1}{n} (- cos (n t) (π - t) - n \cdot \int \frac{1}{n} cos (n t) d t)]_{0}^{π}

\int \frac{1}{n} cos (n t) d t = \frac{1}{n ^{2}} sin (n t)

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

Simplify

[\frac{1}{n} (- cos (n t) (π - t) - n \frac{1}{n ^{2}} sin (n t))]_{0}^{π} : [\frac{1}{n} (- cos (n t) (π - t) - \frac{1}{n} sin (n t))]_{0}^{π}

= [\frac{1}{n} (- cos (n t) (π - t) - \frac{1}{n} sin (n t))]_{0}^{π}

Compute the boundaries:

\frac{π}{n}

= \frac{π}{n}

\int_{0}^{L} cos^{2} (\frac{π x}{2 L}) d x

\int_{1}^{A} \frac{1}{x} d x

\int_{0}^{\frac{7}{5}} \frac{1}{25 x ^{2} + 49} d x

\int_{0}^{π} x (7 + sin (x)) d x

\int_{0}^{1} (4 x + 1)^{2} d x

What is the integral from 0 to pi of (pi-t)sin(nt) ?
The integral from 0 to pi of (pi-t)sin(nt) is pi/n