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

The integral from 0 to pi of xsin((nx)/2) is (4sin((pin)/2)-2pincos((pin)/2))/(n^2)

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{n} (\frac{4}{n} sin (\frac{πn}{2}) - 2 π cos (\frac{πn}{2}))

= \frac{1}{n} (\frac{4}{n} sin (\frac{πn}{2}) - 2 π cos (\frac{πn}{2}))

Simplify

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

\int_{- 1}^{2} (x^{2}) d x

\int_{0}^{1} x^{2} cos (nπ x) d x

\int_{0}^{1} x^{3} 1 + x^{2} d x

\int_{1}^{3} x^{2} + 2 d x

\int_{a}^{b} cos (x) d x

What is the integral from 0 to pi of xsin((nx)/2) ?
The integral from 0 to pi of xsin((nx)/2) is (4sin((pin)/2)-2pincos((pin)/2))/(n^2)