What is the integral from 0 to 1 of cos(3x-3) ?

The integral from 0 to 1 of cos(3x-3) is 1/3 sin(3)

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integral from 0 to 1 of cos(3x-3)

\int_{0}^{1} cos (3 x - 3) d x

\frac{1}{3} sin (3)

Decimal

0.04704 \dots

Solution steps

\int_{0}^{1} cos (3 x - 3) d x

Apply u-substitution

: \int_{- 3}^{0} cos (u) \frac{1}{3} d u

= \int_{- 3}^{0} cos (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{- 3}^{0} cos (u) d u

Use the common integral:

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

= \frac{1}{3} [sin (u)]_{- 3}^{0}

Compute the boundaries:

sin (3)

= \frac{1}{3} sin (3)

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\int_{4}^{7} 3 + 2 ln (t + 5) d t

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What is the integral from 0 to 1 of cos(3x-3) ?
The integral from 0 to 1 of cos(3x-3) is 1/3 sin(3)