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

The integral from-pi to pi of xsin(kx) is -(2picos(pik))/k+(2sin(pik))/(k^2)

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integral from-pi to pi of xsin(kx)

\int_{- π}^{π} x sin (k x) d x

- \frac{2 π cos ( πk )}{k} + \frac{2 sin ( πk )}{k ^{2}}

Solution steps

\int_{- π}^{π} x sin (k x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- 2 \cdot \frac{π}{k} cos (πk) + \frac{2}{k ^{2}} sin (πk)

= - 2 \cdot \frac{π}{k} cos (πk) + \frac{2}{k ^{2}} sin (πk)

Simplify

= - \frac{2 π cos ( πk )}{k} + \frac{2 sin ( πk )}{k ^{2}}

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What is the integral from-pi to pi of xsin(kx) ?
The integral from-pi to pi of xsin(kx) is -(2picos(pik))/k+(2sin(pik))/(k^2)