What is the integral from-5 to 5 of (e^x-e^{-x})^2 ?

The integral from-5 to 5 of (e^x-e^{-x})^2 is (e^{20}-1)/(e^{10)}-20

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integral from-5 to 5 of (e^x-e^{-x})^2

\int_{- 5}^{5} (e^{x} - e^{- x})^{2} d x

\frac{e ^{20} - 1}{e ^{10}} - 20

Decimal

22006.46574 \dots

Solution steps

\int_{- 5}^{5} (e^{x} - e^{- x})^{2} d x

Rewrite using trig identities

= \int_{- 5}^{5} (2 sinh (x))^{2} d x

Simplify

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

= \int_{- 5}^{5} 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_{- 5}^{5} sinh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{- 5}^{5} \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_{- 5}^{5} - 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_{- 5}^{5} 1 d x + \int_{- 5}^{5} cosh (2 x) d x)

\int_{- 5}^{5} 1 d x = 10

\int_{- 5}^{5} cosh (2 x) d x = \frac{e ^{20} - 1}{2 e ^{10}}

= 4 \cdot \frac{1}{2} (- 10 + \frac{e ^{20} - 1}{2 e ^{10}})

Simplify

4 \cdot \frac{1}{2} (- 10 + \frac{e ^{20} - 1}{2 e ^{10}}) : \frac{e ^{20} - 1}{e ^{10}} - 20

= \frac{e ^{20} - 1}{e ^{10}} - 20

x \to - \infty lim (- \frac{1}{0})

n = 0 \sum \infty 3 - (0.5)^{n}

\int_{1}^{\infty} \frac{1}{x ^{6}} d x

\frac{1}{( 2 s + 4 )}

\int (4 x + 3)^{5} d x

What is the integral from-5 to 5 of (e^x-e^{-x})^2 ?
The integral from-5 to 5 of (e^x-e^{-x})^2 is (e^{20}-1)/(e^{10)}-20