What is the integral from-8 to 8 of (e^x-e^{-x})^2 ?

The integral from-8 to 8 of (e^x-e^{-x})^2 is (e^{32}-1)/(e^{16)}-32

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-8 to 8 of (e^x-e^{-x})^2

\int_{- 8}^{8} (e^{x} - e^{- x})^{2} d x

\frac{e ^{32} - 1}{e ^{16}} - 32

Decimal Notation

8886078.52050 \dots

Solution steps

\int_{- 8}^{8} (e^{x} - e^{- x})^{2} d x

Rewrite using trig identities

= \int_{- 8}^{8} (2 sinh (x))^{2} d x

Simplify

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

= \int_{- 8}^{8} 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_{- 8}^{8} sinh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{- 8}^{8} \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_{- 8}^{8} - 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_{- 8}^{8} 1 d x + \int_{- 8}^{8} cosh (2 x) d x)

\int_{- 8}^{8} 1 d x = 16

\int_{- 8}^{8} cosh (2 x) d x = \frac{e ^{32} - 1}{2 e ^{16}}

= 4 \cdot \frac{1}{2} (- 16 + \frac{e ^{32} - 1}{2 e ^{16}})

Simplify

4 \cdot \frac{1}{2} (- 16 + \frac{e ^{32} - 1}{2 e ^{16}}) : \frac{e ^{32} - 1}{e ^{16}} - 32

= \frac{e ^{32} - 1}{e ^{16}} - 32

\int_{0}^{2} x + 1 d x

\frac{\partial}{\partial x} (4 x^{\frac{2}{3}} y^{\frac{1}{3}})

t (e^{t} + e^{2 t})^{2}

\frac{d}{d x} (0.5 e^{\frac{x}{3}} - sin (x))

\int_{0}^{4} x - 2 x d x

What is the integral from-8 to 8 of (e^x-e^{-x})^2 ?
The integral from-8 to 8 of (e^x-e^{-x})^2 is (e^{32}-1)/(e^{16)}-32