What is the integral from 1 to 2 of 2cos(3x) ?

The integral from 1 to 2 of 2cos(3x) is 2/3 (sin(6)-sin(3))

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integral from 1 to 2 of 2cos(3x)

\int_{1}^{2} 2 cos (3 x) d x

\frac{2}{3} (sin (6) - sin (3))

Decimal Notation

- 0.28035 \dots

Solution steps

\int_{1}^{2} 2 cos (3 x) d x

Take the constant out:

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

= 2 \cdot \int_{1}^{2} cos (3 x) d x

Apply u-substitution

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

= 2 \cdot \int_{3}^{6} 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

= 2 \cdot \frac{1}{3} \cdot \int_{3}^{6} cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{3} [sin (u)]_{3}^{6}

Simplify

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

= \frac{2}{3} [sin (u)]_{3}^{6}

Compute the boundaries:

sin (6) - sin (3)

= \frac{2}{3} (sin (6) - sin (3))

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What is the integral from 1 to 2 of 2cos(3x) ?
The integral from 1 to 2 of 2cos(3x) is 2/3 (sin(6)-sin(3))