What is the integral from 1 to 2 of xcos(npix) ?

The integral from 1 to 2 of xcos(npix) is (2pinsin(2pin)+1-(-1)^n)/(pi^2n^2)

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integral from 1 to 2 of xcos(npix)

\int_{1}^{2} x cos (nπ x) d x

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

Solution steps

\int_{1}^{2} x cos (nπ x) d x

Apply u-substitution

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

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

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{πn}^{2 πn}

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

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

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

Simplify

= \frac{1}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{πn}^{2 πn}

Compute the boundaries:

2 πn sin (2 πn) + 1 - (- 1)^{n}

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

Simplify

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

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