What is the integral from 0 to 1/2 of xsin(npix) ?

The integral from 0 to 1/2 of xsin(npix) is (2sin((pin)/2)-pincos((pin)/2))/(2pi^2n^2)

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integral from 0 to 1/2 of xsin(npix)

\int_{0}^{\frac{1}{2}} x sin (nπ x) d x

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

Solution steps

\int_{0}^{\frac{1}{2}} x sin (nπ x) d x

Apply Integration By Parts

: [\frac{1}{πn} (- x cos (πn x) - πn \cdot \int - \frac{1}{πn} cos (πn x) d x)]_{0}^{\frac{1}{2}}

= [\frac{1}{πn} (- x cos (πn x) - πn \cdot \int - \frac{1}{πn} cos (πn x) d x)]_{0}^{\frac{1}{2}}

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

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

Simplify

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

= [\frac{1}{πn} (- x cos (πn x) + \frac{1}{πn} sin (πn x))]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{πn} (\frac{1}{πn} sin (\frac{π}{2} n) - \frac{1}{2} cos (\frac{π}{2} n))

= \frac{1}{πn} (\frac{1}{πn} sin (\frac{π}{2} n) - \frac{1}{2} cos (\frac{π}{2} n))

Simplify

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

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What is the integral from 0 to 1/2 of xsin(npix) ?
The integral from 0 to 1/2 of xsin(npix) is (2sin((pin)/2)-pincos((pin)/2))/(2pi^2n^2)