What is the integral from 0 to 1 of 8cos(2x) ?

The integral from 0 to 1 of 8cos(2x) is 4sin(2)

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integral from 0 to 1 of 8cos(2x)

\int_{0}^{1} 8 cos (2 x) d x

4 sin (2)

Decimal Notation

3.63718 \dots

Solution steps

\int_{0}^{1} 8 cos (2 x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} cos (2 x) d x

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{2} cos (u) d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} [sin (u)]_{0}^{2}

Simplify

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

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

Compute the boundaries:

sin (2)

= 4 sin (2)

\int_{0}^{9} \frac{1}{9 - x} d x

\frac{d}{d x} (4 x^{\frac{5}{4}} + 10 x^{\frac{3}{2}} + 10 x)

\frac{d}{d x} ((x - 4) \cdot a x^{2} + ln (\frac{x}{4}) \cdot b x^{4})

\int cos (8 t) d t

\int \frac{x ^{2} + 2 x - 1}{x ( x + 1 ) ( x - 1 )} d x

What is the integral from 0 to 1 of 8cos(2x) ?
The integral from 0 to 1 of 8cos(2x) is 4sin(2)