What is the integral from 0 to 1 of sin(npit) ?

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

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integral from 0 to 1 of sin(npit)

\int_{0}^{1} sin (nπ t) d t

\frac{- ( - 1 ) ^{n} + 1}{πn}

Solution steps

\int_{0}^{1} sin (nπ t) d t

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{πn} \cdot \int_{0}^{πn} sin (u) d u

Use the common integral:

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

= \frac{1}{πn} [- cos (u)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= \frac{1}{πn} (- (- 1)^{n} + 1)

Simplify

= \frac{- ( - 1 ) ^{n} + 1}{πn}

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