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

The integral from 0 to 2pi of xsin((nx)/2) is -(4pi(-1)^n)/n

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

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

- \frac{4 π ( - 1 ) ^{n}}{n}

Solution steps

\int_{0}^{2 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}^{2 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}^{2 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}^{2 π}

Simplify

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

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

Compute the boundaries:

- \frac{4 π ( - 1 ) ^{n}}{n}

= - \frac{4 π ( - 1 ) ^{n}}{n}

\int_{1}^{2} (4 x - 1) d x

\int_{- 1}^{1} (x^{3} - 3 x) d x

\int_{3}^{5} (4 x^{3} + 5) d x

\int_{\frac{1}{4}}^{\frac{1}{2}} (\frac{π}{x ^{2}}) d x

\int_{- 3}^{- 1} \frac{2}{3 x} d x

What is the integral from 0 to 2pi of xsin((nx)/2) ?
The integral from 0 to 2pi of xsin((nx)/2) is -(4pi(-1)^n)/n