What is the integral from 0 to 2 of xsin((npix)/4) ?

The integral from 0 to 2 of xsin((npix)/4) is (16sin((pin)/2)-8pincos((pin)/2))/(pi^2n^2)

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integral from 0 to 2 of xsin((npix)/4)

\int_{0}^{2} x sin (\frac{nπ x}{4}) d x

\frac{16 sin ( \frac{πn}{2} ) - 8 πn cos ( \frac{πn}{2} )}{π ^{2} n ^{2}}

Solution steps

\int_{0}^{2} x sin (\frac{nπ x}{4}) d x

Apply u-substitution

: \int_{0}^{nπ 2} \frac{u sin ( \frac{u}{4} )}{π ^{2} n ^{2}} d u

= \int_{0}^{nπ 2} \frac{u sin ( \frac{u}{4} )}{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_{0}^{nπ 2} u sin (\frac{u}{4}) d u

Apply Integration By Parts

: [- 4 u cos (\frac{u}{4}) - \int - 4 cos (\frac{u}{4}) d u]_{0}^{2 πn}

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

\int - 4 cos (\frac{u}{4}) d u = - 16 sin (\frac{u}{4})

= \frac{1}{π ^{2} n ^{2}} [- 4 u cos (\frac{u}{4}) - (- 16 sin (\frac{u}{4}))]_{0}^{2 πn}

Simplify

= \frac{1}{π ^{2} n ^{2}} [- 4 u cos (\frac{u}{4}) + 16 sin (\frac{u}{4})]_{0}^{2 πn}

Compute the boundaries:

16 sin (\frac{πn}{2}) - 8 πn cos (\frac{πn}{2})

= \frac{1}{π ^{2} n ^{2}} (16 sin (\frac{πn}{2}) - 8 πn cos (\frac{πn}{2}))

Simplify

= \frac{16 sin ( \frac{πn}{2} ) - 8 πn cos ( \frac{πn}{2} )}{π ^{2} n ^{2}}

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What is the integral from 0 to 2 of xsin((npix)/4) ?
The integral from 0 to 2 of xsin((npix)/4) is (16sin((pin)/2)-8pincos((pin)/2))/(pi^2n^2)