What is the integral from 0 to 1 of 9cos(2x) ?

The integral from 0 to 1 of 9cos(2x) is 9/2 sin(2)

integral from 0 to 1 of 9cos(2x)

\int_{0}^{1} 9 cos (2 x) d x

\frac{9}{2} sin (2)

Decimal

4.09183 \dots

Solution steps

\int_{0}^{1} 9 cos (2 x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{1} cos (2 x) d x

Apply u-substitution

= 9 \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

= 9 \cdot \frac{1}{2} \cdot \int_{0}^{2} cos (u) d u

Use the common integral:

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

= 9 \cdot \frac{1}{2} [sin (u)]_{0}^{2}

Simplify

9 \cdot \frac{1}{2} [sin (u)]_{0}^{2} : \frac{9}{2} [sin (u)]_{0}^{2}

= \frac{9}{2} [sin (u)]_{0}^{2}

Compute the boundaries:

sin (2)

= \frac{9}{2} sin (2)

\frac{d}{d x} (ln (3 - x^{2}))

x \to \infty lim (3 x + 4)

n = 0 \sum \infty (- 1)^{n} (- 1)^{n}

\frac{\partial}{\partial y} (y t sec^{2} (y t))

t an g e n t \frac{x - 3}{9 x - 2}

What is the integral from 0 to 1 of 9cos(2x) ?
The integral from 0 to 1 of 9cos(2x) is 9/2 sin(2)