What is the integral from 0 to 2pi of θsin(nθ) ?

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

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integral from 0 to 2pi of θsin(nθ)

\int_{0}^{2 π} θ sin (n θ) d θ

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

Solution steps

\int_{0}^{2 π} θ sin (n θ) d θ

Apply Integration By Parts

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

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

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

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

Simplify

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

= [\frac{1}{n} (- θ cos (n θ) + \frac{1}{n} sin (n θ))]_{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}}

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