What is the integral from 0 to 1/2 of sin(npix) ?

The integral from 0 to 1/2 of sin(npix) is (-cos((pin)/2)+1)/(pin)

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integral from 0 to 1/2 of sin(npix)

\int_{0}^{\frac{1}{2}} sin (nπ x) d x

\frac{- cos ( \frac{πn}{2} ) + 1}{πn}

Solution steps

\int_{0}^{\frac{1}{2}} sin (nπ x) d x

Apply u-substitution

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

= \int_{0}^{\frac{πn}{2}} \frac{sin ( u )}{πn} d u

Take the constant out:

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

= \frac{1}{πn} \cdot \int_{0}^{\frac{πn}{2}} sin (u) d u

Use the common integral:

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

= \frac{1}{πn} [- cos (u)]_{0}^{\frac{πn}{2}}

Compute the boundaries:

- cos (\frac{πn}{2}) + 1

= \frac{1}{πn} (- cos (\frac{πn}{2}) + 1)

Simplify

= \frac{- cos ( \frac{πn}{2} ) + 1}{πn}

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What is the integral from 0 to 1/2 of sin(npix) ?
The integral from 0 to 1/2 of sin(npix) is (-cos((pin)/2)+1)/(pin)