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

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

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integral of sin((npix)/a)

\int sin (\frac{nπ x}{a}) d x

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

Solution steps

\int sin (\frac{nπ x}{a}) d x

Apply u-substitution

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

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

Apply u-substitution

: \int sin (v) a d v

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

Use the common integral:

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

= \frac{1}{πn} a (- cos (v))

Substitute back

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

Simplify

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

Add a constant to the solution

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

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(0.9) (4)

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