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

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

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integral of sin(t-x)

\int sin (t - x) d x

cos (t - x) + C

Solution steps

\int sin (t - x) d x

Apply u-substitution

: \int - sin (u) d u

= \int - sin (u) d u

Take the constant out:

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

= - \int sin (u) d u

Use the common integral:

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

= - (- cos (u))

Substitute back

u = t - x

= - (- cos (t - x))

Simplify

= cos (t - x)

Add a constant to the solution

= cos (t - x) + C

x \to 0 + lim (coth (x))

\int - 4 x d x

\int \frac{2 x}{1 + x} d x

x \to 2 π lim (x (csc (x)))

\int_{0}^{1} (2 x^{2} + 1) d x