What is the integral from 0 to pi of-sin((nx)/2) ?

The integral from 0 to pi of-sin((nx)/2) is -(2(-cos((pin)/2)+1))/(n)

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integral from 0 to pi of-sin((nx)/2)

\int_{0}^{π} - sin (\frac{n x}{2}) d x

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

Solution steps

\int_{0}^{π} - sin (\frac{n x}{2}) d x

Take the constant out:

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

= - \int_{0}^{π} sin (\frac{n x}{2}) d x

Apply u-substitution

: \int_{0}^{\frac{nπ}{2}} \frac{2 sin ( u )}{n} d u

= - \int_{0}^{\frac{nπ}{2}} \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_{0}^{\frac{nπ}{2}} sin (u) d u

Use the common integral:

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

= - \frac{2}{n} [- cos (u)]_{0}^{\frac{nπ}{2}}

Compute the boundaries:

- cos (\frac{πn}{2}) + 1

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

Simplify

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

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What is the integral from 0 to pi of-sin((nx)/2) ?
The integral from 0 to pi of-sin((nx)/2) is -(2(-cos((pin)/2)+1))/(n)