What is the integral of sin((npix)/L) ?

The integral of sin((npix)/L) is -L 1/(pin)cos((pinx)/L)+C

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integral of sin((npix)/L)

\int sin (\frac{nπ x}{L}) d x

- L \frac{1}{πn} cos (\frac{πn x}{L}) + C

Solution steps

\int sin (\frac{nπ x}{L}) d x

Apply u-substitution

: \int \frac{sin ( \frac{u}{L} )}{πn} d u

= \int \frac{sin ( \frac{u}{L} )}{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 sin (\frac{u}{L}) d u

Apply u-substitution

: \int sin (v) L d v

= \frac{1}{πn} \cdot \int sin (v) L d v

Take the constant out:

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

= \frac{1}{πn} L \cdot \int sin (v) d v

Use the common integral:

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

= \frac{1}{πn} L (- cos (v))

Substitute back

= \frac{1}{πn} L (- cos (\frac{n x π}{L}))

Simplify

= - L \frac{1}{πn} cos (\frac{πn x}{L})

Add a constant to the solution

= - L \frac{1}{πn} cos (\frac{πn x}{L}) + C

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What is the integral of sin((npix)/L) ?
The integral of sin((npix)/L) is -L 1/(pin)cos((pinx)/L)+C