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

The integral from pi/2 to pi of xsin(nx) is -(pi(-1)^n)/n-(-pincos((pin)/2)+2sin((pin)/2))/(2n^2)

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

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What is the integral from pi/2 to pi of xsin(nx) ?
The integral from pi/2 to pi of xsin(nx) is -(pi(-1)^n)/n-(-pincos((pin)/2)+2sin((pin)/2))/(2n^2)