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

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

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

\int_{- 1}^{0} (1 - x) cos (nπ x) d x

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

Solution steps

\int_{- 1}^{0} (1 - x) cos (nπ x) d x

Expand

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

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

\int_{- 1}^{0} cos (πn x) d x = 0

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

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

Simplify

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

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