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

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

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integral from-pi to 0 of sin(nx)

\int_{- π}^{0} sin (n x) d x

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

Solution steps

\int_{- π}^{0} sin (n x) d x

Apply u-substitution

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

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

Use the common integral:

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

= \frac{1}{n} [- cos (u)]_{- πn}^{0}

Compute the boundaries:

- 1 + (- 1)^{n}

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

Simplify

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

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