What is the integral from-2 to 0 of xcos((npix)/2) ?

The integral from-2 to 0 of xcos((npix)/2) is (4(1-(-1)^n))/(pi^2n^2)

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integral from-2 to 0 of xcos((npix)/2)

\int_{- 2}^{0} x cos (\frac{nπ x}{2}) d x

\frac{4 ( 1 - ( - 1 ) ^{n} )}{π ^{2} n ^{2}}

Solution steps

\int_{- 2}^{0} x cos (\frac{nπ x}{2}) d x

Apply u-substitution

: \int_{- 2 πn}^{0} \frac{u cos ( \frac{u}{2} )}{π ^{2} n ^{2}} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int_{- πn}^{0} 4 v cos (v) d v

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{- πn}^{0} 4 v cos (v) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{π ^{2} n ^{2}} \cdot 4 \cdot \int_{- πn}^{0} v cos (v) d v

Apply Integration By Parts

: [v sin (v) - \int sin (v) d v]_{- πn}^{0}

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - \int sin (v) d v]_{- πn}^{0}

\int sin (v) d v = - cos (v)

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{- πn}^{0}

Simplify

\frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{- πn}^{0} : \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{- πn}^{0}

= \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{- πn}^{0}

Compute the boundaries:

1 - (- 1)^{n}

= \frac{4}{π ^{2} n ^{2}} (1 - (- 1)^{n})

Simplify

= \frac{4 ( 1 - ( - 1 ) ^{n} )}{π ^{2} n ^{2}}

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What is the integral from-2 to 0 of xcos((npix)/2) ?
The integral from-2 to 0 of xcos((npix)/2) is (4(1-(-1)^n))/(pi^2n^2)