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

The integral from 0 to pi of (pi-x)sin(nx) is pi/n

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integral from 0 to pi of (pi-x)sin(nx)

\int_{0}^{π} (π - x) sin (n x) d x

\frac{π}{n}

Solution steps

\int_{0}^{π} (π - x) sin (n x) d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{n}

= \frac{π}{n}

What is the integral from 0 to pi of (pi-x)sin(nx) ?
The integral from 0 to pi of (pi-x)sin(nx) is pi/n