What is the integral from 0 to 1 of 6cos(2x) ?

The integral from 0 to 1 of 6cos(2x) is 3sin(2)

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integral from 0 to 1 of 6cos(2x)

\int_{0}^{1} 6 cos (2 x) d x

3 sin (2)

Decimal Notation

2.72789 \dots

Solution steps

\int_{0}^{1} 6 cos (2 x) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} cos (2 x) d x

Apply u-substitution

: \int_{0}^{2} cos (u) \frac{1}{2} d u

= 6 \cdot \int_{0}^{2} cos (u) \frac{1}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{2} cos (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} [sin (u)]_{0}^{2}

Simplify

6 \cdot \frac{1}{2} [sin (u)]_{0}^{2} : 3 [sin (u)]_{0}^{2}

= 3 [sin (u)]_{0}^{2}

Compute the boundaries:

sin (2)

= 3 sin (2)

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What is the integral from 0 to 1 of 6cos(2x) ?
The integral from 0 to 1 of 6cos(2x) is 3sin(2)