What is the integral from-4 to 4 of (e^x-e^{-x})^2 ?

The integral from-4 to 4 of (e^x-e^{-x})^2 is (e^{16}-1)/(e^8)-16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-4 to 4 of (e^x-e^{-x})^2

\int_{- 4}^{4} (e^{x} - e^{- x})^{2} d x

\frac{e ^{16} - 1}{e ^{8}} - 16

Decimal Notation

2964.95765 \dots

Solution steps

\int_{- 4}^{4} (e^{x} - e^{- x})^{2} d x

Rewrite using trig identities

= \int_{- 4}^{4} (2 sinh (x))^{2} d x

Simplify

(2 sinh (x))^{2} : 4 sinh^{2} (x)

= \int_{- 4}^{4} 4 sinh^{2} (x) d x

Take the constant out:

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

= 4 \cdot \int_{- 4}^{4} sinh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{- 4}^{4} \frac{- 1 + cosh ( 2 x )}{2} d x

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{- 4}^{4} - 1 + cosh (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{2} (- \int_{- 4}^{4} 1 d x + \int_{- 4}^{4} cosh (2 x) d x)

\int_{- 4}^{4} 1 d x = 8

\int_{- 4}^{4} cosh (2 x) d x = \frac{e ^{16} - 1}{2 e ^{8}}

= 4 \cdot \frac{1}{2} (- 8 + \frac{e ^{16} - 1}{2 e ^{8}})

Simplify

4 \cdot \frac{1}{2} (- 8 + \frac{e ^{16} - 1}{2 e ^{8}}) : \frac{e ^{16} - 1}{e ^{8}} - 16

= \frac{e ^{16} - 1}{e ^{8}} - 16

\frac{\partial}{\partial y} ((x y)^{\frac{1}{2}})

\int \frac{( 8 x ^{2} + 3 x - 6 )}{( x ^{2} - 4 ) ( x + 2 )} d x

\frac{\partial}{\partial y} (2 arctan (\frac{x}{y}))

2 x^{e}

\frac{d y}{d x} = 4 y^{2} x

What is the integral from-4 to 4 of (e^x-e^{-x})^2 ?
The integral from-4 to 4 of (e^x-e^{-x})^2 is (e^{16}-1)/(e^8)-16