What is the integral from 0 to 1 of 8cos(3x) ?

The integral from 0 to 1 of 8cos(3x) is 8/3 sin(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 8cos(3x)

\int_{0}^{1} 8 cos (3 x) d x

\frac{8}{3} sin (3)

Decimal Notation

0.37632 \dots

Solution steps

\int_{0}^{1} 8 cos (3 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 (3 x) d x

Apply u-substitution

: \int_{0}^{3} cos (u) \frac{1}{3} d u

= 8 \cdot \int_{0}^{3} 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

= 8 \cdot \frac{1}{3} \cdot \int_{0}^{3} cos (u) d u

Use the common integral:

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

= 8 \cdot \frac{1}{3} [sin (u)]_{0}^{3}

Simplify

8 \cdot \frac{1}{3} [sin (u)]_{0}^{3} : \frac{8}{3} [sin (u)]_{0}^{3}

= \frac{8}{3} [sin (u)]_{0}^{3}

Compute the boundaries:

sin (3)

= \frac{8}{3} sin (3)

f (x) = 4 cos^{2} (x)

\frac{\partial}{\partial x} (2 x y - x^{2} y)

\int 32 x^{3} - 15 x^{2} + 10 x d x

x + \frac{1}{y} = 2

\int \frac{2 x - 3}{( x - 1 ) ^{2}} d x

What is the integral from 0 to 1 of 8cos(3x) ?
The integral from 0 to 1 of 8cos(3x) is 8/3 sin(3)