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

The integral from 0 to x of ysin(x-t) is -y(-1+cos(x))

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

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

- y (- 1 + cos (x))

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= y \cdot \int_{x}^{0} - sin (u) d u

Take the constant out:

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

= y (- \int_{x}^{0} sin (u) d u)

Use the common integral:

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

= y (- [- cos (u)]_{x}^{0})

Simplify

= - y [- cos (u)]_{x}^{0}

Compute the boundaries:

- 1 + cos (x)

= - y (- 1 + cos (x))

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What is the integral from 0 to x of ysin(x-t) ?
The integral from 0 to x of ysin(x-t) is -y(-1+cos(x))