What is the integral from 0 to pi of xcos((nx)/2) ?

The integral from 0 to pi of xcos((nx)/2) is (2pinsin((pin)/2)+4cos((pin)/2)-4)/(n^2)

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integral from 0 to pi of xcos((nx)/2)

\int_{0}^{π} x cos (\frac{n x}{2}) d x

\frac{2 πn sin ( \frac{πn}{2} ) + 4 cos ( \frac{πn}{2} ) - 4}{n ^{2}}

Solution steps

\int_{0}^{π} x cos (\frac{n x}{2}) d x

Apply Integration By Parts

: [\frac{1}{n} (2 x sin (\frac{n x}{2}) - n \cdot \int \frac{2}{n} sin (\frac{n x}{2}) d x)]_{0}^{π}

= [\frac{1}{n} (2 x sin (\frac{n x}{2}) - n \cdot \int \frac{2}{n} sin (\frac{n x}{2}) d x)]_{0}^{π}

\int \frac{2}{n} sin (\frac{n x}{2}) d x = - \frac{4}{n ^{2}} cos (\frac{n x}{2})

= [\frac{1}{n} (2 x sin (\frac{n x}{2}) - n (- \frac{4}{n ^{2}} cos (\frac{n x}{2})))]_{0}^{π}

Simplify

[\frac{1}{n} (2 x sin (\frac{n x}{2}) - n (- \frac{4}{n ^{2}} cos (\frac{n x}{2})))]_{0}^{π} : [\frac{1}{n} (2 x sin (\frac{n x}{2}) + \frac{4}{n} cos (\frac{n x}{2}))]_{0}^{π}

= [\frac{1}{n} (2 x sin (\frac{n x}{2}) + \frac{4}{n} cos (\frac{n x}{2}))]_{0}^{π}

Compute the boundaries:

\frac{1}{n} (2 π sin (\frac{nπ}{2}) + \frac{4}{n} cos (\frac{nπ}{2})) - \frac{4}{n ^{2}}

= \frac{1}{n} (2 π sin (\frac{nπ}{2}) + \frac{4}{n} cos (\frac{nπ}{2})) - \frac{4}{n ^{2}}

Simplify

= \frac{2 πn sin ( \frac{πn}{2} ) + 4 cos ( \frac{πn}{2} ) - 4}{n ^{2}}

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What is the integral from 0 to pi of xcos((nx)/2) ?
The integral from 0 to pi of xcos((nx)/2) is (2pinsin((pin)/2)+4cos((pin)/2)-4)/(n^2)