Question

If $ \int\limits_{\log 2 }^{x} \frac{dy}{\sqrt{e^y - 1}} = \frac{\pi}{6}$ then $x$ is equal to

• A. $\log_e 4$
• B. $\log_e 2$
• C. $5$
• D. $2$

Answer 1

Answer: (A)

We have, $ \int\limits_{\log 2 }^{x} \frac{dy}{\sqrt{e^y - 1}} = \frac{\pi}{6}$
Put $e^y = \sec^2 \theta , \, \, i.e., y = \log (\sec^2 \theta )$
$ \Rightarrow \, \, dy = \frac{1}{\sec^2 \theta} 2 \, \sec \theta \, \sec \theta \, \tan \theta \, d \theta = 2 \tan \theta \, d \theta $
If $ y = \log 2,$ then $\log 2 = \log(\sec^2 \theta)$
$ \Rightarrow \, \sec \theta = \sqrt{2}$
$ \Rightarrow \, \theta = \frac{\pi}{4}$
If $y = x$, then $x = \log \, \sec^2 \theta \, \Rightarrow e^x = \sec^2 \theta$
$ \Rightarrow \, \, \sqrt{e^x} = \sec \theta \, \, \Rightarrow \theta = \sec^{-1} \sqrt{e^x}$
$ \therefore \, \, \int\limits_{\log2}^{x} \frac{dy}{\sqrt{e^{y} -1}} = \int\limits_{\frac{\pi}{4}}^{\sec^{-1}\sqrt{e^{x}}} \frac{2 \tan \theta d\theta}{\sqrt{\sec^{2} \theta -1 }}$
$ \frac{\pi}{6} = \int\limits_{\frac{\pi }{4}}^{\sec ^{-1}\sqrt{e^{x} 2} } \frac{2 \tan \theta}{\tan \theta} d\theta = 2\left[\theta\right]_{\pi /4}^{\sec^{-1\left(e^{x/ 2}\right)}} $
$\Rightarrow \frac{\pi}{6} = 2 \left[\sec ^{-1\left(e^{x 2}\right)} - \frac{\pi}{4}\right] $
$\Rightarrow \sec ^{-1\left(e^{x 2}\right)}- \frac{\pi}{4} $
$ \Rightarrow e^{x/ 2} =\sec \frac{\pi}{3} = 2 $
Taking logarithms on both sides, we get
$\frac{x}{2} =\log 2 \Rightarrow x = 2 \log2 =\log 2^{2} =\log 4$
$ x =\log 4$