What is the integral from-pi to pi of xsin(nx) ?

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

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integral from-pi to pi of xsin(nx)

\int_{- π}^{π} x sin (n x) d x

- \frac{2 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{- π}^{π} x sin (n x) d x

Apply Integration By Parts

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

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

Simplify

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

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

Compute the boundaries:

- 2 (- 1)^{n} \frac{π}{n}

= - 2 (- 1)^{n} \frac{π}{n}

Simplify

= - \frac{2 π ( - 1 ) ^{n}}{n}

\frac{d}{d x} (\frac{x + 1}{f ( x ) + 4})

\int (\frac{1}{x ^{5}} - x^{5} - \frac{1}{4}) d x

\int 2 y e^{- y^{2}} d y

\frac{1}{s ^{2}} e^{- s}

x \to 2 lim (2 x + 1)

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