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

The integral from 0 to 2 of xsin(nx) is (sin(2n)}{n^2}-\frac{2cos(2n))/n

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integral from 0 to 2 of xsin(nx)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 0 to 2 of xsin(nx) ?
The integral from 0 to 2 of xsin(nx) is (sin(2n)}{n^2}-\frac{2cos(2n))/n