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

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

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integral from 0 to a of sin((npix)/a)

\int_{0}^{a} sin (\frac{nπ x}{a}) d x

\frac{a ( - ( - 1 ) ^{n} + 1 )}{πn}

Solution steps

\int_{0}^{a} sin (\frac{nπ x}{a}) d x

Apply u-substitution

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

= \int_{0}^{nπa} \frac{sin ( \frac{u}{a} )}{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}^{nπa} sin (\frac{u}{a}) d u

Apply u-substitution

: \int_{0}^{πn} sin (v) a d v

= \frac{1}{πn} \cdot \int_{0}^{πn} sin (v) a d v

Take the constant out:

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

= \frac{1}{πn} a \cdot \int_{0}^{πn} sin (v) d v

Use the common integral:

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

= \frac{1}{πn} a [- cos (v)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= \frac{1}{πn} a (- (- 1)^{n} + 1)

Simplify

= \frac{a ( - ( - 1 ) ^{n} + 1 )}{πn}

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