What is the integral from 0 to 1 of xsin((npi)/2 x) ?

The integral from 0 to 1 of xsin((npi)/2 x) is (2(2sin((pin)/2)-pincos((pin)/2)))/(pi^2n^2)

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integral from 0 to 1 of xsin((npi)/2 x)

\int_{0}^{1} x sin (\frac{nπ}{2} x) d x

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

Solution steps

\int_{0}^{1} x sin (\frac{nπ}{2} x) d x

\frac{nπ}{2} = nπ \frac{1}{2}

= \int_{0}^{1} x sin (nπ \frac{1}{2} x) d x

Apply u-substitution

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

= \int_{0}^{\frac{πn}{2}} \frac{4 u sin ( 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{4}{π ^{2} n ^{2}} \cdot \int_{0}^{\frac{πn}{2}} u sin (u) d u

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

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

= \frac{4}{π ^{2} n ^{2}} (sin (\frac{πn}{2}) - \frac{πn}{2} cos (\frac{πn}{2}))

Simplify

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

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What is the integral from 0 to 1 of xsin((npi)/2 x) ?
The integral from 0 to 1 of xsin((npi)/2 x) is (2(2sin((pin)/2)-pincos((pin)/2)))/(pi^2n^2)