What is the integral from 0 to t of sin(w(t-x)) ?

The integral from 0 to t of sin(w(t-x)) is -(-1+cos(wt))/w

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integral from 0 to t of sin(w(t-x))

\int_{0}^{t} sin (w (t - x)) d x

- \frac{- 1 + cos ( wt )}{w}

Solution steps

\int_{0}^{t} sin (w (t - x)) d x

Apply u-substitution

: \int_{t}^{0} - sin (w u) d u

= \int_{t}^{0} - sin (w u) d u

Take the constant out:

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

= - \int_{t}^{0} sin (w u) d u

Apply u-substitution

: \int_{wt}^{0} \frac{sin ( v )}{w} d v

= - \int_{wt}^{0} \frac{sin ( v )}{w} d v

Take the constant out:

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

= - \frac{1}{w} \cdot \int_{wt}^{0} sin (v) d v

Use the common integral:

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

= - \frac{1}{w} [- cos (v)]_{wt}^{0}

Compute the boundaries:

- 1 + cos (wt)

= - \frac{1}{w} (- 1 + cos (wt))

Simplify

= - \frac{- 1 + cos ( wt )}{w}

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What is the integral from 0 to t of sin(w(t-x)) ?
The integral from 0 to t of sin(w(t-x)) is -(-1+cos(wt))/w