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

The integral from 0 to pi of (pi-x)*cos(nx) is (-(-1)^n+1)/(n^2)

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integral from 0 to pi of (pi-x)*cos(nx)

\int_{0}^{π} (π - x) \cdot cos (n x) d x

\frac{- ( - 1 ) ^{n} + 1}{n ^{2}}

Solution steps

\int_{0}^{π} (π - x) cos (n x) d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

- \frac{( - 1 ) ^{n}}{n ^{2}} + \frac{1}{n ^{2}}

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

Simplify

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

What is the integral from 0 to pi of (pi-x)*cos(nx) ?
The integral from 0 to pi of (pi-x)*cos(nx) is (-(-1)^n+1)/(n^2)