What is the integral of xsin((npi)/2 x) ?

The integral of xsin((npi)/2 x) is 4/(pi^2n^2)(-pi/2 nxcos(pi/2 nx)+sin(pi/2 nx))+C

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integral of xsin((npi)/2 x)

\int x sin (\frac{nπ}{2} x) d x

\frac{4}{π ^{2} n ^{2}} (- \frac{π}{2} n x cos (\frac{π}{2} n x) + sin (\frac{π}{2} n x)) + C

Solution steps

\int x sin (\frac{nπ}{2} x) d x

\frac{nπ}{2} = nπ \frac{1}{2}

= \int x sin (nπ \frac{1}{2} x) d x

Apply u-substitution

: \int \frac{4 u sin ( u )}{π ^{2} n ^{2}} d u

= \int \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 u sin (u) d u

Apply Integration By Parts

: - u cos (u) - \int - cos (u) d u

= \frac{4}{π ^{2} n ^{2}} (- u cos (u) - \int - cos (u) d u)

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

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

Substitute back

u = nπ \frac{1}{2} x

= \frac{4}{π ^{2} n ^{2}} (- nπ \frac{1}{2} x cos (nπ \frac{1}{2} x) - (- sin (nπ \frac{1}{2} x)))

Simplify

\frac{4}{π ^{2} n ^{2}} (- nπ \frac{1}{2} x cos (nπ \frac{1}{2} x) - (- sin (nπ \frac{1}{2} x))) : \frac{4}{π ^{2} n ^{2}} (- \frac{π}{2} n x cos (\frac{π}{2} n x) + sin (\frac{π}{2} n x))

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

Add a constant to the solution

= \frac{4}{π ^{2} n ^{2}} (- \frac{π}{2} n x cos (\frac{π}{2} n x) + sin (\frac{π}{2} n x)) + C

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What is the integral of xsin((npi)/2 x) ?
The integral of xsin((npi)/2 x) is 4/(pi^2n^2)(-pi/2 nxcos(pi/2 nx)+sin(pi/2 nx))+C