What is the integral from 1 to 2 of cos((nxpi)/2) ?

The integral from 1 to 2 of cos((nxpi)/2) is -(2sin((pin)/2))/(pin)

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integral from 1 to 2 of cos((nxpi)/2)

\int_{1}^{2} cos (\frac{n x π}{2}) d x

- \frac{2 sin ( \frac{πn}{2} )}{πn}

Solution steps

\int_{1}^{2} cos (\frac{n x π}{2}) d x

Apply u-substitution

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

= \int_{\frac{πn}{2}}^{πn} \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

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

Use the common integral:

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

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

Compute the boundaries:

- sin (\frac{πn}{2})

= \frac{2}{πn} (- sin (\frac{πn}{2}))

Simplify

= - \frac{2 sin ( \frac{πn}{2} )}{πn}

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