What is the integral from-1 to 0 of-2cos((npix)/2) ?

The integral from-1 to 0 of-2cos((npix)/2) is -(4sin((pin)/2))/(pin)

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integral from-1 to 0 of-2cos((npix)/2)

\int_{- 1}^{0} - 2 cos (\frac{nπ x}{2}) d x

- \frac{4 sin ( \frac{πn}{2} )}{πn}

Solution steps

\int_{- 1}^{0} - 2 cos (\frac{nπ x}{2}) d x

Take the constant out:

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

= - 2 \cdot \int_{- 1}^{0} cos (\frac{nπ x}{2}) d x

Apply u-substitution

: \int_{- \frac{πn}{2}}^{0} \frac{2 cos ( u )}{πn} d u

= - 2 \cdot \int_{- \frac{πn}{2}}^{0} \frac{2 cos ( u )}{πn} d u

Take the constant out:

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

= - 2 \cdot \frac{2}{πn} \cdot \int_{- \frac{πn}{2}}^{0} cos (u) d u

Use the common integral:

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

= - 2 \cdot \frac{2}{πn} [sin (u)]_{- \frac{πn}{2}}^{0}

Simplify

- 2 \cdot \frac{2}{πn} [sin (u)]_{- \frac{πn}{2}}^{0} : - \frac{4}{πn} [sin (u)]_{- \frac{πn}{2}}^{0}

= - \frac{4}{πn} [sin (u)]_{- \frac{πn}{2}}^{0}

Compute the boundaries:

sin (\frac{πn}{2})

= - \frac{4}{πn} sin (\frac{πn}{2})

Simplify

= - \frac{4 sin ( \frac{πn}{2} )}{πn}

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What is the integral from-1 to 0 of-2cos((npix)/2) ?
The integral from-1 to 0 of-2cos((npix)/2) is -(4sin((pin)/2))/(pin)