What is the integral from 0 to t of sin(a(t-u)) ?

The integral from 0 to t of sin(a(t-u)) is -(-1+cos(at))/a

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integral from 0 to t of sin(a(t-u))

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

- \frac{- 1 + cos ( a t )}{a}

Solution steps

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

Apply u-substitution

: \int_{t}^{0} - sin (a v) d v

= \int_{t}^{0} - sin (a v) d v

Take the constant out:

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

= - \int_{t}^{0} sin (a v) d v

Apply u-substitution

: \int_{a t}^{0} \frac{sin ( w )}{a} d w

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

- 1 + cos (a t)

= - \frac{1}{a} (- 1 + cos (a t))

Simplify

= - \frac{- 1 + cos ( a t )}{a}

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What is the integral from 0 to t of sin(a(t-u)) ?
The integral from 0 to t of sin(a(t-u)) is -(-1+cos(at))/a