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

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

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

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

1 - cos (x)

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Compute the boundaries:

- 1 + cos (x)

= - (- 1 + cos (x))

Simplify

= 1 - cos (x)

\int_{0}^{π} x cos (5 x) d x

\int_{0}^{1} 3 x^{2} - 3 x d x

\int_{1}^{e} \frac{( 1 + ln ( x ) ) ^{4}}{x} d x

\int_{- 1}^{1} e^{3 x} d x

\int_{0}^{1} ln (1 + y) d y

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