What is the integral from 0 to 2pi of x^2sin(nx) ?

The integral from 0 to 2pi of x^2sin(nx) is -(4pi^2cos(2pin))/n+(4pisin(2pin))/(n^2)

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integral from 0 to 2pi of x^2sin(nx)

\int_{0}^{2 π} x^{2} sin (n x) d x

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

Solution steps

\int_{0}^{2 π} x^{2} sin (n x) d x

Apply Integration By Parts

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

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

\int - \frac{2}{n} x cos (n x) d x = - \frac{2}{n ^{3}} (n x sin (n x) + cos (n x))

= [\frac{1}{n} (- x^{2} cos (n x) - n (- \frac{2}{n ^{3}} (n x sin (n x) + cos (n x))))]_{0}^{2 π}

Simplify

[\frac{1}{n} (- x^{2} cos (n x) - n (- \frac{2}{n ^{3}} (n x sin (n x) + cos (n x))))]_{0}^{2 π} : [\frac{1}{n} (- x^{2} cos (n x) + \frac{2}{n ^{2}} (n x sin (n x) + cos (n x)))]_{0}^{2 π}

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

Compute the boundaries:

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

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

Simplify

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

\frac{7}{x - 2}

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\frac{d}{d x} ((2 x^{3} + 5)^{10})

What is the integral from 0 to 2pi of x^2sin(nx) ?
The integral from 0 to 2pi of x^2sin(nx) is -(4pi^2cos(2pin))/n+(4pisin(2pin))/(n^2)