What is the integral of sin((npix)/4) ?

The integral of sin((npix)/4) is -4/(pin)cos((pinx)/4)+C

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integral of sin((npix)/4)

\int sin (\frac{nπ x}{4}) d x

- \frac{4}{πn} cos (\frac{πn x}{4}) + C

Solution steps

\int sin (\frac{nπ x}{4}) d x

Apply u-substitution

: \int \frac{4 sin ( u )}{πn} d u

= \int \frac{4 sin ( u )}{πn} d u

Take the constant out:

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

= \frac{4}{πn} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{4}{πn} (- cos (u))

Substitute back

u = \frac{n x π}{4}

= \frac{4}{πn} (- cos (\frac{n x π}{4}))

Simplify

= - \frac{4}{πn} cos (\frac{πn x}{4})

Add a constant to the solution

= - \frac{4}{πn} cos (\frac{πn x}{4}) + C

\frac{d}{d x} (\frac{2 + sin ( x )}{2 x + cos ( x )})

y = ∣ x ∣, y = x^{2} - 1, x = - 1, x = 1

5 t \frac{d y}{d t} - 5 y = t

\frac{\partial}{\partial y} (\frac{y}{x ^{2} + y ^{2}})

\frac{\partial}{\partial x} (x^{3} z + yz)

What is the integral of sin((npix)/4) ?
The integral of sin((npix)/4) is -4/(pin)cos((pinx)/4)+C