What is the integral from 0 to 1 of sin^2(kpix) ?

The integral from 0 to 1 of sin^2(kpix) is 1/2-1/(4pik)sin(2pik)

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integral from 0 to 1 of sin^2(kpix)

\int_{0}^{1} sin^{2} (kπ x) d x

\frac{1}{2} - \frac{1}{4 πk} sin (2 πk)

Solution steps

\int_{0}^{1} sin^{2} (kπ x) d x

Rewrite using trig identities

= \int_{0}^{1} \frac{1 - cos ( 2 kπ x )}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} 1 - cos (2 kπ x) d x

Apply u-substitution

: \int_{0}^{2 πk} \frac{1 - cos ( u )}{2 πk} d u

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

Take the constant out:

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

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

\int_{0}^{2 πk} 1 d u = 2 πk

\int_{0}^{2 πk} cos (u) d u = sin (2 πk)

= \frac{1}{2} \cdot \frac{1}{2 πk} (2 πk - sin (2 πk))

Simplify

\frac{1}{2} \cdot \frac{1}{2 πk} (2 πk - sin (2 πk)) : \frac{1}{2} - \frac{1}{4 πk} sin (2 πk)

= \frac{1}{2} - \frac{1}{4 πk} sin (2 πk)

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What is the integral from 0 to 1 of sin^2(kpix) ?
The integral from 0 to 1 of sin^2(kpix) is 1/2-1/(4pik)sin(2pik)