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

The integral from 0 to pi of sin(kx) is (-cos(pik)+1)/k

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integral from 0 to pi of sin(kx)

\int_{0}^{π} sin (k x) d x

\frac{- cos ( πk ) + 1}{k}

Solution steps

\int_{0}^{π} sin (k x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{k} [- cos (u)]_{0}^{kπ}

Compute the boundaries:

- cos (πk) + 1

= \frac{1}{k} (- cos (πk) + 1)

Simplify

= \frac{- cos ( πk ) + 1}{k}

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What is the integral from 0 to pi of sin(kx) ?
The integral from 0 to pi of sin(kx) is (-cos(pik)+1)/k