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

The integral from 0 to 1/2 of xsin(pikx) is (2sin((pik)/2)-pikcos((pik)/2))/(2pi^2k^2)

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{πk} (\frac{1}{πk} sin (\frac{π}{2} k) - \frac{1}{2} cos (\frac{π}{2} k))

= \frac{1}{πk} (\frac{1}{πk} sin (\frac{π}{2} k) - \frac{1}{2} cos (\frac{π}{2} k))

Simplify

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

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What is the integral from 0 to 1/2 of xsin(pikx) ?
The integral from 0 to 1/2 of xsin(pikx) is (2sin((pik)/2)-pikcos((pik)/2))/(2pi^2k^2)