What is the integral from 0 to a/2 of sin^2((pix)/a) ?

The integral from 0 to a/2 of sin^2((pix)/a) is a/4

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integral from 0 to a/2 of sin^2((pix)/a)

\int_{0}^{\frac{a}{2}} sin^{2} (\frac{π x}{a}) d x

\frac{a}{4}

Solution steps

\int_{0}^{\frac{a}{2}} sin^{2} (\frac{π x}{a}) d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{a}{π} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u

Rewrite using trig identities

= \frac{a}{π} \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{a}{π} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{a}{π} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= \frac{a}{π} \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

\frac{a}{π} \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{a}{4}

= \frac{a}{4}

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What is the integral from 0 to a/2 of sin^2((pix)/a) ?
The integral from 0 to a/2 of sin^2((pix)/a) is a/4