What is the integral of sin(npix) ?

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

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integral of sin(npix)

\int sin (nπ x) d x

- \frac{1}{πn} cos (πn x) + C

Solution steps

\int sin (nπ x) d x

Apply u-substitution

: \int \frac{sin ( u )}{πn} d u

= \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 \int sin (u) d u

Use the common integral:

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

= \frac{1}{πn} (- cos (u))

Substitute back

u = nπ x

= \frac{1}{πn} (- cos (nπ x))

Simplify

= - \frac{1}{πn} cos (πn x)

Add a constant to the solution

= - \frac{1}{πn} cos (πn x) + C

\frac{d}{d x} ((x + 1) (x + 3))

\frac{( x + 1 )}{( 5 - x )}

t \to 1 lim (\frac{\frac{1}{t}}{t - 1})

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

\frac{d}{d x} (sin (2 x + π))

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