What is the integral from-pi to pi of (1+x)sin(nx) ?

The integral from-pi to pi of (1+x)sin(nx) is -(2pi(-1)^n)/n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-pi to pi of (1+x)sin(nx)

\int_{- π}^{π} (1 + x) sin (n x) d x

- \frac{2 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{π} (1 + x) sin (n x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- (- 1)^{n} \frac{π + 1}{n} + (- 1)^{n} \frac{- π + 1}{n}

= - (- 1)^{n} \frac{π + 1}{n} + (- 1)^{n} \frac{- π + 1}{n}

Simplify

= - \frac{2 π ( - 1 ) ^{n}}{n}

\int_{0}^{5} (2 e^{x}) d x

\int_{0}^{3} cos (\frac{nπ x}{3}) d x

\int_{1}^{2} (\frac{5 - x}{x ^{3}}) d x

\int_{a}^{\infty} \frac{1}{x ^{p}} d x

\int_{1}^{3} (2 x + 5) d x

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