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

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

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integral from 0 to pi of sin(x-kx)

\int_{0}^{π} sin (x - k x) d x

\frac{- cos ( π - πk ) + 1}{1 - k}

Solution steps

\int_{0}^{π} sin (x - k x) d x

Apply u-substitution

: \int_{0}^{π - kπ} \frac{sin ( u )}{1 - k} d u

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{1 - k} [- cos (u)]_{0}^{π - kπ}

Compute the boundaries:

- cos (π - πk) + 1

= \frac{1}{1 - k} (- cos (π - πk) + 1)

Simplify

= \frac{- cos ( π - πk ) + 1}{1 - k}

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