What is the integral of sin((npix)/2) ?

The integral of sin((npix)/2) is -2/(pin)cos((pin)/2 x)+C

integral of sin((npix)/2)

\int sin (\frac{nπ x}{2}) d x

- \frac{2}{πn} cos (\frac{πn}{2} x) + C

Solution steps

\int sin (\frac{nπ x}{2}) d x

\frac{nπ x}{2} = \frac{nπ}{2} x

= \int sin (\frac{nπ}{2} x) d x

Apply u-substitution

= \int \frac{2 sin ( u )}{πn} d u

Take the constant out:

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

= \frac{2}{πn} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{2}{πn} (- cos (u))

Substitute back

u = \frac{nπ}{2} x

= \frac{2}{πn} (- cos (\frac{nπ}{2} x))

Simplify

= - \frac{2}{πn} cos (\frac{πn}{2} x)

Add a constant to the solution

= - \frac{2}{πn} cos (\frac{πn}{2} x) + C

\int e^{c o s (16 t)} sin (16 t) d t

\int y^{2} cos (x) d x

\frac{\partial}{\partial x} (x + \frac{y}{z})

y^{^{''}} - 8 y^{^{'}} + 17 y = 0

3 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

What is the integral of sin((npix)/2) ?
The integral of sin((npix)/2) is -2/(pin)cos((pin)/2 x)+C