What is the integral from 0 to t of cos(2x-t) ?

The integral from 0 to t of cos(2x-t) is sin(t)

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integral from 0 to t of cos(2x-t)

\int_{0}^{t} cos (2 x - t) d x

sin (t)

Solution steps

\int_{0}^{t} cos (2 x - t) d x

Apply u-substitution

: \int_{- t}^{t} cos (u) \frac{1}{2} d u

= \int_{- t}^{t} cos (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- t}^{t} cos (u) d u

Use the common integral:

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

= \frac{1}{2} [sin (u)]_{- t}^{t}

Compute the boundaries:

2 sin (t)

= \frac{1}{2} \cdot 2 sin (t)

Simplify

= sin (t)

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What is the integral from 0 to t of cos(2x-t) ?
The integral from 0 to t of cos(2x-t) is sin(t)