What is the integral from 0 to pi of sin(nt) ?

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

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integral from 0 to pi of sin(nt)

\int_{0}^{π} sin (n t) d t

\frac{- ( - 1 ) ^{n} + 1}{n}

Solution steps

\int_{0}^{π} sin (n t) d t

Apply u-substitution

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

= \int_{0}^{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_{0}^{nπ} sin (u) d u

Use the common integral:

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

= \frac{1}{n} [- cos (u)]_{0}^{nπ}

Compute the boundaries:

- (- 1)^{n} + 1

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

Simplify

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

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