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

The integral from-pi/2 to pi/2 of xsin(nx) is -(picos((pin)/2))/(n)+(2sin((pin)/2))/(n^2)

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integral from-pi/2 to pi/2 of xsin(nx)

\int_{- \frac{π}{2}}^{\frac{π}{2}} x sin (n x) d x

- \frac{π cos ( \frac{πn}{2} )}{n} + \frac{2 sin ( \frac{πn}{2} )}{n ^{2}}

Solution steps

\int_{- \frac{π}{2}}^{\frac{π}{2}} 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)]_{- \frac{π}{2}}^{\frac{π}{2}}

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

\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)))]_{- \frac{π}{2}}^{\frac{π}{2}}

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= - \frac{π cos ( \frac{πn}{2} )}{n} + \frac{2 sin ( \frac{πn}{2} )}{n ^{2}}

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