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

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

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integral from 0 to 2pi of xsin(nx)

\int_{0}^{2 π} x sin (n x) d x

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

Solution steps

\int_{0}^{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)]_{0}^{2 π}

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{1} 3 x + cos (3 x - 3) d x

\int_{0}^{\frac{π}{2}} x^{2} d x

\int_{4}^{6} (x^{2} + 2 x - 7) d x

\int_{\frac{3 π}{2}}^{π} x cos (x) d x

\int_{4}^{6} (x^{2} + 2 x - 6) d x

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