First slide

Evaluation of definite integrals

Question

$\int_{\log 2}^{k} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6} then k =$

Moderate

A

4
B

log 8
C

log 4
D

None

Solution

$\int_{\log 2}^{k} \frac{e^{x} d x}{e^{x} \cdot \sqrt{e^{x} - 1}} put \sqrt{e^{x} - 1} = t$
$\int_{1}^{\sqrt{e} - 1} \frac{2 t d t}{(t^{2} + 1) t} \begin{aligned} e^{x} - 1 = t^{2} \\ e^{x} d x = 2 t d t \end{aligned}$
$2 \int_{1}^{\sqrt{e^{ε} - 1}} \frac{d t}{1 + t^{2}} = 2 {[\tan^{- 1} t]}_{1}^{\sqrt{e^{t} - 1}} \begin{aligned} L L \\ \begin{matrix} t = \sqrt{2 - 1} \\ t = 1 \end{matrix} \end{aligned} ∣ \begin{aligned} U L \\ t = \sqrt{e^{k} - 1} \end{aligned}$
$\begin{array}{l} \Rightarrow 2 [\tan^{- 1} (\sqrt{e^{k} - 1}) - \frac{π}{4}] = \frac{π}{6} \\ 2 \tan^{- 1} (\sqrt{e^{k} - 1}) = \frac{π}{6} + \frac{π}{2} \\ \tan^{- 1} (\sqrt{e^{k} - 1}) = \frac{π}{3} \\ (\sqrt{e^{k} - 1}) = \sqrt{3} \\ e^{k} = 4 \\ k = \log 4 \end{array}$

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