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

The integral from 0 to t of sin(t-x) is 1-cos(t)

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

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

1 - cos (t)

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

- 1 + cos (t)

= - (- 1 + cos (t))

Simplify

= 1 - cos (t)

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\int_{0}^{π} \frac{1}{2} sin (x^{2}) d x

\int_{- 2}^{3} x^{3} - 6 x d x

\int_{2}^{7} \frac{1}{x + 2 + 1} d x

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