What is the integral of sin(t-u) ?

The integral of sin(t-u) is cos(t-u)+C

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integral of sin(t-u)

\int sin (t - u) d u

cos (t - u) + C

Solution steps

\int sin (t - u) d u

Apply u-substitution

= \int - sin (v) d v

Take the constant out:

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

= - \int sin (v) d v

Use the common integral:

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

= - (- cos (v))

Substitute back

v = t - u

= - (- cos (t - u))

Simplify

= cos (t - u)

Add a constant to the solution

= cos (t - u) + C