What is the integral from 1 to 2 of sin((npix)/2) ?

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

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integral from 1 to 2 of sin((npix)/2)

\int_{1}^{2} sin (\frac{nπ x}{2}) d x

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

Solution steps

\int_{1}^{2} sin (\frac{nπ x}{2}) d x

\frac{nπ x}{2} = \frac{nπ}{2} x

= \int_{1}^{2} sin (\frac{nπ}{2} x) d x

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

- (- 1)^{n} + cos (\frac{πn}{2})

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

Simplify

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

\int_{0}^{1} 3 x^{2} (2 x^{3} - 3)^{2} d x

\frac{d}{d x} (4 (x + 2)^{3})

x \to - 3 lim (- (\frac{1}{x ^{2} - 9}))

\int \frac{3 x ^{3} + 6 x ^{2} + 3}{3 x + 3} d x

x \to \infty lim (0.6 6^{x})

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