What is the integral from 1 to x of cos(t-3) ?

The integral from 1 to x of cos(t-3) is sin(x-3)+sin(2)

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integral from 1 to x of cos(t-3)

\int_{1}^{x} cos (t - 3) d t

sin (x - 3) + sin (2)

Solution steps

\int_{1}^{x} cos (t - 3) d t

Apply u-substitution

: \int_{- 2}^{x - 3} cos (u) d u

= \int_{- 2}^{x - 3} cos (u) d u

Use the common integral:

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

= [sin (u)]_{- 2}^{x - 3}

Compute the boundaries:

sin (x - 3) + sin (2)

= sin (x - 3) + sin (2)

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

\int_{0}^{2} \frac{48}{x ^{2} + 4} d x

\int_{0}^{π} 1 + cos (4 x) d x

\int_{- 3}^{3} 27 - 3 x^{2} d x

\int_{- 2}^{- 1} \frac{5}{2 x} d x

What is the integral from 1 to x of cos(t-3) ?
The integral from 1 to x of cos(t-3) is sin(x-3)+sin(2)