What is the integral of 2/(npi)sin((npix)/2) ?

The integral of 2/(npi)sin((npix)/2) is -4/(pi^2n^2)cos((pin)/2 x)+C

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integral of 2/(npi)sin((npix)/2)

\int \frac{2}{nπ} sin (\frac{nπ x}{2}) d x

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

Solution steps

\int \frac{2}{nπ} sin (\frac{nπ x}{2}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{nπ} \cdot \int sin (\frac{nπ x}{2}) d x

\frac{nπ x}{2} = \frac{nπ}{2} x

= \frac{2}{nπ} \cdot \int sin (\frac{nπ}{2} x) d x

Apply u-substitution

: \int \frac{2 sin ( u )}{πn} d u

= \frac{2}{nπ} \cdot \int \frac{2 sin ( u )}{πn} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{nπ} \cdot \frac{2}{πn} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{2}{nπ} \cdot \frac{2}{πn} (- cos (u))

Substitute back

u = \frac{nπ}{2} x

= \frac{2}{nπ} \cdot \frac{2}{πn} (- cos (\frac{nπ}{2} x))

Simplify

\frac{2}{nπ} \cdot \frac{2}{πn} (- cos (\frac{nπ}{2} x)) : - \frac{4}{π ^{2} n ^{2}} cos (\frac{πn}{2} x)

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

Add a constant to the solution

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

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What is the integral of 2/(npi)sin((npix)/2) ?
The integral of 2/(npi)sin((npix)/2) is -4/(pi^2n^2)cos((pin)/2 x)+C