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

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

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integral from 0 to l of sin^2((npix)/l)

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

\frac{1}{2} (l - \frac{l sin ( 2 πn )}{2 πn})

Solution steps

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

Rewrite using trig identities

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

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} (\int_{0}^{l} 1 d x - \int_{0}^{l} cos (\frac{2 πn x}{l}) d x)

\int_{0}^{l} 1 d x = l

\int_{0}^{l} cos (\frac{2 πn x}{l}) d x = \frac{l sin ( 2 πn )}{2 πn}

= \frac{1}{2} (l - \frac{l sin ( 2 πn )}{2 πn})

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