Question
Let $\left( 2^{1 - a} + 2^{1 + a} \right),f(a),\left( 3^{a} + 3^{- a} \right)$ be a A. P. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_{e}(\alpha - 1)}^{\log_{e}(\alpha)}\mspace{2mu}\frac{dx}{\left( e^{2x} - e^{- 2x} \right)}$ is :
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Solution
Sol. [Key Concept: The A.P. condition gives $f(a) = \frac{1}{2}{(2(2}^{a}{+ 2}^{- a}{) + (3}^{a}{+ 3}^{- a}))$; its minimum is $\alpha = 3$, then integrate with $e^{2x} = t$.]
$$f(a) = \frac{1}{2}\left( 2\left( 2^{a} + 2^{- a} \right) + \left( 3^{a} + 3^{- a} \right) \right)$$
$$\begin{matrix} & a = \frac{1}{2}(2 \times 2 + 2) = 3 \\ & I = \int_{ln(3 - 1)}^{ln3}\mspace{2mu}\mspace{2mu}\frac{e^{2x}}{e^{4x} - 1}dx \end{matrix}$$
put $e^{2x} = t$
$${I = \frac{1}{2}\int_{4}^{9}\mspace{2mu}\frac{dt}{t^{2} - 1} }{I = \frac{1}{4}\left( ln\left( \frac{t - 1}{t + 1} \right) \right)_{4}^{9} }{I = \frac{1}{4}ln\left( \frac{4}{3} \right) }$$
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