What is the integral from 0 to 1 of 9cos(3x) ?

The integral from 0 to 1 of 9cos(3x) is 3sin(3)

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integral from 0 to 1 of 9cos(3x)

\int_{0}^{1} 9 cos (3 x) d x

3 sin (3)

Decimal

0.42336 \dots

Solution steps

\int_{0}^{1} 9 cos (3 x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{1} cos (3 x) d x

Apply u-substitution

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

= 9 \cdot \int_{0}^{3} 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

= 9 \cdot \frac{1}{3} \cdot \int_{0}^{3} cos (u) d u

Use the common integral:

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

= 9 \cdot \frac{1}{3} [sin (u)]_{0}^{3}

Simplify

9 \cdot \frac{1}{3} [sin (u)]_{0}^{3} : 3 [sin (u)]_{0}^{3}

= 3 [sin (u)]_{0}^{3}

Compute the boundaries:

sin (3)

= 3 sin (3)

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What is the integral from 0 to 1 of 9cos(3x) ?
The integral from 0 to 1 of 9cos(3x) is 3sin(3)