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

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

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

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

\frac{2 ( - 1 + ( - 1 ) ^{n} )}{πn}

Solution steps

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

\frac{nπ x}{2} = \frac{nπ}{2} x

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

Apply u-substitution

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

= \int_{- πn}^{0} \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}{πn} \cdot \int_{- πn}^{0} sin (u) d u

Use the common integral:

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

= \frac{2}{πn} [- cos (u)]_{- πn}^{0}

Compute the boundaries:

- 1 + (- 1)^{n}

= \frac{2}{πn} (- 1 + (- 1)^{n})

Simplify

= \frac{2 ( - 1 + ( - 1 ) ^{n} )}{πn}

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