First slide

Evaluation of definite integrals

Question

Solution of the equation $\int_{\log 2}^{x} \frac{d x}{\sqrt{e^{x} - 1}} = \frac{π}{6}$ are

Moderate

A

$x = \log 6$
B

$x = 2 \log 2$
C

$x = 3$
D

$x = 1 / 2$

Solution

Putting $e^{x} - 1 = t^{2}$ , we have $e^{x} d x = 2 t d t$
so $\int_{\log 2}^{x} \frac{d x}{\sqrt{e^{x} - 1}} = \int_{1}^{\sqrt{e^{x} - 1}} \frac{2 t d t}{t (t^{2} + 1)} = 2 \int_{1}^{\sqrt{e^{x} - 1}} \frac{d t}{t^{2} + 1}$
$= 2 (\tan^{- 1} \sqrt{e^{x} - 1} - \frac{π}{4})$
Thus the given equation reduces to
$\tan^{- 1} \sqrt{e^{x} - 1} - \frac{π}{4} = \frac{π}{12} \Rightarrow \sqrt{e^{x} - 1} = \tan \frac{π}{3} = \sqrt{3}$
so $e^{x} = 4 \Rightarrow x = 2 \log 2$

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