Step by step solution to "Evaluate the following integrals. Include absolute values only..."

alesterp

Answered question

2021-10-10

Evaluate the following integrals. Include absolute values only when needed.

\int_{0}^{\ln 2} \frac{e^{3 x} - e^{- 3 x}}{e^{3 x} + e^{- 3 x}}

Benedict

Skilled2021-10-11Added 108 answers

The given integral is

\int_{0}^{\ln 2} \frac{e^{3 x} - e^{- 3 x}}{e^{3 x} + e^{- 3 x}}

Evaluate as follows.

\int_{0}^{\ln 2} \frac{e^{3 x} - e^{- 3 x}}{e^{3 x} + e^{- 3 x}} = \frac{1}{3} \int_{0}^{\ln 2} \frac{3 e^{3 x} - 3 e^{- 3 x}}{e^{3 x} + e^{- 3 x}} d x

= \frac{1}{3} \int_{0}^{\ln 2} \frac{d (e^{3 x} + e^{- 3 x})}{e^{3 x} + e^{- 3 x}}

= \frac{1}{3} {[\ln (e^{3 x} + e^{- 3 x})]}_{0}^{\ln 2}

= \frac{1}{3} [\ln (e^{3 \ln 2} + e^{- 3 \ln 2}) - \ln (e^{0} + e^{0})]

= \frac{1}{3} [\ln (2^{3} + 2^{- 3}) - \ln (1 + 1)]

= \frac{1}{3} \ln (\frac{2^{2} + 2^{- 3}}{2})

Thus,

\int_{0}^{\ln 2} \frac{e^{3 x} - e^{- 3 x}}{e^{3 x} + e^{- 3 x}} = \frac{1}{3} \ln (\frac{2^{2} + 2^{- 3}}{2})

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