What is the integral from-6 to 6 of (e^x-e^{-x})^2 ?

The integral from-6 to 6 of (e^x-e^{-x})^2 is (e^{24}-1)/(e^{12)}-24

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integral from-6 to 6 of (e^x-e^{-x})^2

\int_{- 6}^{6} (e^{x} - e^{- x})^{2} d x

\frac{e ^{24} - 1}{e ^{12}} - 24

Decimal Notation

162730.79141 \dots

Solution steps

\int_{- 6}^{6} (e^{x} - e^{- x})^{2} d x

Rewrite using trig identities

= \int_{- 6}^{6} (2 sinh (x))^{2} d x

Simplify

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

= \int_{- 6}^{6} 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_{- 6}^{6} sinh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{- 6}^{6} \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_{- 6}^{6} - 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_{- 6}^{6} 1 d x + \int_{- 6}^{6} cosh (2 x) d x)

\int_{- 6}^{6} 1 d x = 12

\int_{- 6}^{6} cosh (2 x) d x = \frac{e ^{24} - 1}{2 e ^{12}}

= 4 \cdot \frac{1}{2} (- 12 + \frac{e ^{24} - 1}{2 e ^{12}})

Simplify

4 \cdot \frac{1}{2} (- 12 + \frac{e ^{24} - 1}{2 e ^{12}}) : \frac{e ^{24} - 1}{e ^{12}} - 24

= \frac{e ^{24} - 1}{e ^{12}} - 24

\frac{cos ( t )}{t ^{7}}

\frac{\partial}{\partial b} (\frac{ab}{2 c + 15})

x \to 0 lim (x e^{2 x})

x \to 0 lim (\frac{4}{x csc ( x )})

y = 3 (x^{3} - x), y = 0

What is the integral from-6 to 6 of (e^x-e^{-x})^2 ?
The integral from-6 to 6 of (e^x-e^{-x})^2 is (e^{24}-1)/(e^{12)}-24