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

The integral from 0 to 1 of sech^2(5x) is (e^{10}-1)/(5(e^{10)+1)}

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

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

\frac{e ^{10} - 1}{5 ( e ^{10} + 1 )}

Decimal Notation

0.19998 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{5} sech^{2} (u) \frac{1}{5} d u

= \int_{0}^{5} sech^{2} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int_{0}^{5} sech^{2} (u) d u

Use the common integral:

\int sech^{2} (u) d u = tanh (u)

= \frac{1}{5} [tanh (u)]_{0}^{5}

Compute the boundaries:

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

= \frac{1}{5} \cdot \frac{e ^{10} - 1}{e ^{10} + 1}

Simplify

= \frac{e ^{10} - 1}{5 ( e ^{10} + 1 )}

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What is the integral from 0 to 1 of sech^2(5x) ?
The integral from 0 to 1 of sech^2(5x) is (e^{10}-1)/(5(e^{10)+1)}