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

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

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integral from 0 to 2 of sin((npi)/2 x)

\int_{0}^{2} sin (\frac{nπ}{2} x) d x

\frac{2 ( - ( - 1 ) ^{n} + 1 )}{πn}

Solution steps

\int_{0}^{2} sin (\frac{nπ}{2} x) d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

= \frac{2}{πn} [- cos (u)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= \frac{2}{πn} (- (- 1)^{n} + 1)

Simplify

= \frac{2 ( - ( - 1 ) ^{n} + 1 )}{πn}

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