What is the integral from 0 to 1 of (1-x)cos(npix) ?

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

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integral from 0 to 1 of (1-x)cos(npix)

\int_{0}^{1} (1 - x) cos (nπ x) d x

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

Solution steps

\int_{0}^{1} (1 - x) cos (nπ x) d x

Expand

(1 - x) cos (nπ x) : cos (πn x) - x cos (πn x)

= \int_{0}^{1} cos (πn x) - x cos (πn x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} cos (πn x) d x - \int_{0}^{1} x cos (πn x) d x

\int_{0}^{1} cos (πn x) d x = 0

\int_{0}^{1} x cos (πn x) d x = \frac{( - 1 ) ^{n} - 1}{π ^{2} n ^{2}}

= 0 - \frac{( - 1 ) ^{n} - 1}{π ^{2} n ^{2}}

Simplify

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

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