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

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

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

\int_{\frac{π}{2}}^{π} sin (n x) d x

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

Solution steps

\int_{\frac{π}{2}}^{π} sin (n x) d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{n} \cdot \int_{\frac{πn}{2}}^{nπ} sin (u) d u

Use the common integral:

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

= \frac{1}{n} [- cos (u)]_{\frac{πn}{2}}^{nπ}

Compute the boundaries:

- (- 1)^{n} + cos (\frac{πn}{2})

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

Simplify

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

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