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

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

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integral from 0 to 1 of 1/2 cosh(2x)

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

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

Decimal Notation

0.90671 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{2} cosh (u) \frac{1}{2} d u

= \frac{1}{2} \cdot \int_{0}^{2} cosh (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{2} cosh (u) d u

Use the common integral:

\int cosh (u) d u = sinh (u)

= \frac{1}{2} \cdot \frac{1}{2} [sinh (u)]_{0}^{2}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [sinh (u)]_{0}^{2} : \frac{1}{4} [sinh (u)]_{0}^{2}

= \frac{1}{4} [sinh (u)]_{0}^{2}

Compute the boundaries:

\frac{e ^{4} - 1}{2 e ^{2}}

= \frac{1}{4} \cdot \frac{e ^{4} - 1}{2 e ^{2}}

Simplify

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

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