What is the integral from 0 to 1 of 5(cosh(x))^2 ?

The integral from 0 to 1 of 5(cosh(x))^2 is (5(4e^2+e^4-1))/(8e^2)

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integral from 0 to 1 of 5(cosh(x))^2

\int_{0}^{1} 5 (cosh (x))^{2} d x

\frac{5 ( 4 e ^{2} + e ^{4} - 1 )}{8 e ^{2}}

Decimal

7.03357 \dots

Solution steps

\int_{0}^{1} 5 (cosh (x))^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{0}^{1} (cosh (x))^{2} d x

Simplify

= 5 \cdot \int_{0}^{1} cosh^{2} (x) d x

Rewrite using trig identities

= 5 \cdot \int_{0}^{1} \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

= 5 \cdot \frac{1}{2} \cdot \int_{0}^{1} 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

= 5 \cdot \frac{1}{2} (\int_{0}^{1} 1 d x + \int_{0}^{1} cosh (2 x) d x)

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

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

= 5 \cdot \frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}})

Simplify

5 \cdot \frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}}) : \frac{5 ( 4 e ^{2} + e ^{4} - 1 )}{8 e ^{2}}

= \frac{5 ( 4 e ^{2} + e ^{4} - 1 )}{8 e ^{2}}

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What is the integral from 0 to 1 of 5(cosh(x))^2 ?
The integral from 0 to 1 of 5(cosh(x))^2 is (5(4e^2+e^4-1))/(8e^2)