What is the integral from pi to 2pi of sin((nx)/2) ?

The integral from pi to 2pi of sin((nx)/2) is (2(-(-1)^n+cos((pin)/2)))/(n)

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integral from pi to 2pi of sin((nx)/2)

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

\frac{2 ( - ( - 1 ) ^{n} + cos ( \frac{πn}{2} ) )}{n}

Solution steps

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

Apply u-substitution

: \int_{\frac{nπ}{2}}^{πn} \frac{2 sin ( u )}{n} d u

= \int_{\frac{nπ}{2}}^{πn} \frac{2 sin ( u )}{n} d u

Take the constant out:

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

= \frac{2}{n} \cdot \int_{\frac{nπ}{2}}^{πn} sin (u) d u

Use the common integral:

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

= \frac{2}{n} [- cos (u)]_{\frac{nπ}{2}}^{πn}

Compute the boundaries:

- (- 1)^{n} + cos (\frac{πn}{2})

= \frac{2}{n} (- (- 1)^{n} + cos (\frac{πn}{2}))

Simplify

= \frac{2 ( - ( - 1 ) ^{n} + cos ( \frac{πn}{2} ) )}{n}

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What is the integral from pi to 2pi of sin((nx)/2) ?
The integral from pi to 2pi of sin((nx)/2) is (2(-(-1)^n+cos((pin)/2)))/(n)