What is the integral from 0 to 1/2 of ysin((npiy)/2) ?

The integral from 0 to 1/2 of ysin((npiy)/2) is (4sin((pin)/4)-pincos((pin)/4))/(pi^2n^2)

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integral from 0 to 1/2 of ysin((npiy)/2)

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

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

Solution steps

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

Apply u-substitution

= \int_{0}^{\frac{πn}{4}} \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}{4}} u sin (u) d u

Apply Integration By Parts

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

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

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

Simplify

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

Compute the boundaries:

sin (\frac{πn}{4}) - \frac{πn}{4} cos (\frac{πn}{4})

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

Simplify

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

What is the integral from 0 to 1/2 of ysin((npiy)/2) ?
The integral from 0 to 1/2 of ysin((npiy)/2) is (4sin((pin)/4)-pincos((pin)/4))/(pi^2n^2)