What is the integral from 0 to L/2 of (2x)/L sin((npix)/L) ?

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

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integral from 0 to L/2 of (2x)/L sin((npix)/L)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{L} \cdot \int_{0}^{\frac{L}{2}} x sin (\frac{n x π}{L}) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

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