What is the integral of sin((npi)/l x) ?

The integral of sin((npi)/l x) is -l 1/(pin)cos((pinx)/l)+C

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integral of sin((npi)/l x)

\int sin (\frac{nπ}{l} x) d x

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

Solution steps

\int sin (\frac{nπ}{l} x) d x

\frac{nπ}{l} = nπ \frac{1}{l}

= \int sin (nπ \frac{1}{l} x) d x

Multiply

nπ \frac{1}{l} x : \frac{πn x}{l}

= \int sin (\frac{πn 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{πn x}{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((npi)/l x) ?
The integral of sin((npi)/l x) is -l 1/(pin)cos((pinx)/l)+C