What is the integral of 1/(pin)sin(npix) ?

The integral of 1/(pin)sin(npix) is -1/(pi^2n^2)cos(pinx)+C

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integral of 1/(pin)sin(npix)

\int \frac{1}{πn} sin (nπ x) d x

- \frac{1}{π ^{2} n ^{2}} cos (πn x) + C

Solution steps

\int \frac{1}{πn} sin (nπ x) d x

Take the constant out:

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

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

Apply u-substitution

: \int \frac{sin ( u )}{πn} d u

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

Use the common integral:

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

= \frac{1}{πn} \cdot \frac{1}{πn} (- cos (u))

Substitute back

u = nπ x

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

Simplify

\frac{1}{πn} \cdot \frac{1}{πn} (- cos (nπ x)) : - \frac{1}{π ^{2} n ^{2}} cos (πn x)

= - \frac{1}{π ^{2} n ^{2}} cos (πn x)

Add a constant to the solution

= - \frac{1}{π ^{2} n ^{2}} cos (πn x) + C

f (x) = x^{2} (3 + x)^{- 5}

\frac{d}{d t} (e^{t} (cos (t) - sin (t)))

f (x) = ln (25 sin (2) (x))

y^{^{''}} + \frac{1}{4} y = 0, y (π) = 1, y^{^{'}} (π) = - 3

\frac{d}{d x} ((\frac{x - 1}{x ^{2} + x + 1})^{4})

What is the integral of 1/(pin)sin(npix) ?
The integral of 1/(pin)sin(npix) is -1/(pi^2n^2)cos(pinx)+C