What is the integral from 0 to 2 of 2sin((npix)/2) ?

The integral from 0 to 2 of 2sin((npix)/2) is (4(-(-1)^n+1))/(pin)

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integral from 0 to 2 of 2sin((npix)/2)

\int_{0}^{2} 2 sin (\frac{nπ x}{2}) d x

\frac{4 ( - ( - 1 ) ^{n} + 1 )}{πn}

Solution steps

\int_{0}^{2} 2 sin (\frac{nπ x}{2}) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} sin (\frac{nπ x}{2}) d x

\frac{nπ x}{2} = \frac{nπ}{2} x

= 2 \cdot \int_{0}^{2} sin (\frac{nπ}{2} x) d x

Apply u-substitution

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

= 2 \cdot \int_{0}^{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

= 2 \cdot \frac{2}{πn} \cdot \int_{0}^{πn} sin (u) d u

Use the common integral:

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

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

Simplify

2 \cdot \frac{2}{πn} [- cos (u)]_{0}^{πn} : \frac{4}{πn} [- cos (u)]_{0}^{πn}

= \frac{4}{πn} [- cos (u)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= \frac{4}{πn} (- (- 1)^{n} + 1)

Simplify

= \frac{4 ( - ( - 1 ) ^{n} + 1 )}{πn}

\int_{0}^{4} \frac{1}{1 + u} d u

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\int_{- 1}^{1} - 2 x^{2} + 2 d x

\int_{0}^{1} sin (x^{4}) d x

\int_{- 1}^{1} x^{2} 4 + x^{2} d x

What is the integral from 0 to 2 of 2sin((npix)/2) ?
The integral from 0 to 2 of 2sin((npix)/2) is (4(-(-1)^n+1))/(pin)