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

f (v) = (v^{- 3} + 3 v^{- 2})^{3}

f^{^{'}} (x) = 7 x^{4}

x \to - 3 lim (x^{2} + 7 x - 6)

x \to 9 + lim (\frac{6}{x - 9})

f (x) = x 5 - x

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