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

The integral from 0 to t of ycos(t-x) is ysin(t)

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integral from 0 to t of ycos(t-x)

\int_{0}^{t} y cos (t - x) d x

y sin (t)

Solution steps

\int_{0}^{t} y cos (t - x) d x

Take the constant out:

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

= y \cdot \int_{0}^{t} cos (t - x) d x

Apply u-substitution

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

= y \cdot \int_{t}^{0} - cos (u) d u

Take the constant out:

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

= y (- \int_{t}^{0} cos (u) d u)

Use the common integral:

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

= y (- [sin (u)]_{t}^{0})

Simplify

= - y [sin (u)]_{t}^{0}

Compute the boundaries:

- sin (t)

= - y (- sin (t))

Simplify

= y sin (t)

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What is the integral from 0 to t of ycos(t-x) ?
The integral from 0 to t of ycos(t-x) is ysin(t)