What is the integral from-7 to 7 of (e^x-e^{-x})^2 ?

The integral from-7 to 7 of (e^x-e^{-x})^2 is (e^{28}-1)/(e^{14)}-28

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integral from-7 to 7 of (e^x-e^{-x})^2

\int_{- 7}^{7} (e^{x} - e^{- x})^{2} d x

\frac{e ^{28} - 1}{e ^{14}} - 28

Decimal Notation

1202576.28416 \dots

Solution steps

\int_{- 7}^{7} (e^{x} - e^{- x})^{2} d x

Rewrite using trig identities

= \int_{- 7}^{7} (2 sinh (x))^{2} d x

Simplify

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

= \int_{- 7}^{7} 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_{- 7}^{7} sinh^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int_{- 7}^{7} \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_{- 7}^{7} - 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_{- 7}^{7} 1 d x + \int_{- 7}^{7} cosh (2 x) d x)

\int_{- 7}^{7} 1 d x = 14

\int_{- 7}^{7} cosh (2 x) d x = \frac{e ^{28} - 1}{2 e ^{14}}

= 4 \cdot \frac{1}{2} (- 14 + \frac{e ^{28} - 1}{2 e ^{14}})

Simplify

4 \cdot \frac{1}{2} (- 14 + \frac{e ^{28} - 1}{2 e ^{14}}) : \frac{e ^{28} - 1}{e ^{14}} - 28

= \frac{e ^{28} - 1}{e ^{14}} - 28

\frac{d}{d x} (\frac{x + x}{x ^{2}})

\int_{0}^{1} 3 ln (x) d x

\int cos (x) sin (n x) d x

\frac{d}{d x} ((cos (3 x)) \cdot 3)

f (x) = - x

What is the integral from-7 to 7 of (e^x-e^{-x})^2 ?
The integral from-7 to 7 of (e^x-e^{-x})^2 is (e^{28}-1)/(e^{14)}-28