What is the integral from 0 to 2 of (e^x-e^{-x})^2 ?

The integral from 0 to 2 of (e^x-e^{-x})^2 is 2((e^8-1)/(4e^4)-2)

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integral from 0 to 2 of (e^x-e^{-x})^2

\int_{0}^{2} (e^{x} - e^{- x})^{2} d x

2 (\frac{e ^{8} - 1}{4 e ^{4}} - 2)

Decimal Notation

23.28991 \dots

Solution steps

\int_{0}^{2} (e^{x} - e^{- x})^{2} d x

Rewrite using trig identities

= \int_{0}^{2} (2 sinh (x))^{2} d x

Simplify

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

= \int_{0}^{2} 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_{0}^{2} sinh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{0}^{2} \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_{0}^{2} - 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_{0}^{2} 1 d x + \int_{0}^{2} cosh (2 x) d x)

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

\int_{0}^{2} cosh (2 x) d x = \frac{e ^{8} - 1}{4 e ^{4}}

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

Simplify

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

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

\int x - 3 d x

y^{^{''}} - 2 y^{^{'}} + y = e^{x} arctan (x)

\int x sin (14 x) d x

x \to - \frac{π}{2} - lim (sec (x))

f (x) = 4 sin (x), (\frac{π}{6}, 2)

What is the integral from 0 to 2 of (e^x-e^{-x})^2 ?
The integral from 0 to 2 of (e^x-e^{-x})^2 is 2((e^8-1)/(4e^4)-2)