What is the integral from 2 to 3 of 6sin(3x) ?

The integral from 2 to 3 of 6sin(3x) is 2(-cos(9)+cos(6))

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integral from 2 to 3 of 6sin(3x)

\int_{2}^{3} 6 sin (3 x) d x

2 (- cos (9) + cos (6))

Decimal

3.74260 \dots

Solution steps

\int_{2}^{3} 6 sin (3 x) d x

Take the constant out:

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

= 6 \cdot \int_{2}^{3} sin (3 x) d x

Apply u-substitution

: \int_{6}^{9} sin (u) \frac{1}{3} d u

= 6 \cdot \int_{6}^{9} sin (u) \frac{1}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int_{6}^{9} sin (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} [- cos (u)]_{6}^{9}

Simplify

6 \cdot \frac{1}{3} [- cos (u)]_{6}^{9} : 2 [- cos (u)]_{6}^{9}

= 2 [- cos (u)]_{6}^{9}

Compute the boundaries:

- cos (9) + cos (6)

= 2 (- cos (9) + cos (6))

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What is the integral from 2 to 3 of 6sin(3x) ?
The integral from 2 to 3 of 6sin(3x) is 2(-cos(9)+cos(6))