What is the integral from 0 to 1 of 6cos(4x) ?

The integral from 0 to 1 of 6cos(4x) is 3/2 sin(4)

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integral from 0 to 1 of 6cos(4x)

\int_{0}^{1} 6 cos (4 x) d x

\frac{3}{2} sin (4)

Decimal

- 1.13520 \dots

Solution steps

\int_{0}^{1} 6 cos (4 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 (4 x) d x

Apply u-substitution

= 6 \cdot \int_{0}^{4} cos (u) \frac{1}{4} d u

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int_{0}^{4} cos (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{4} [sin (u)]_{0}^{4}

Simplify

6 \cdot \frac{1}{4} [sin (u)]_{0}^{4} : \frac{3}{2} [sin (u)]_{0}^{4}

= \frac{3}{2} [sin (u)]_{0}^{4}

Compute the boundaries:

sin (4)

= \frac{3}{2} sin (4)

What is the integral from 0 to 1 of 6cos(4x) ?
The integral from 0 to 1 of 6cos(4x) is 3/2 sin(4)