Question

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

• A. ${\log }_{e}4$
• B. ${\log }_{e}2$
• C. $5$
• D. $2$

Answer 1

Answer: (A)

We have, $\underset{\log 2}{\overset{x}{\int }}\frac{dy}{\sqrt{e^y-1}}=\frac{\pi }{6}$
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 $x=\log {\sec }^2\theta \Rightarrow e^x={\sec }^2\theta$
$\Rightarrow \sqrt{e^x}=\sec \theta \Rightarrow \theta ={\sec }^{-1}\sqrt{e^x}$
$\therefore \underset{\log 2}{\overset{x}{\int }}\frac{dy}{\sqrt{e^y-1}}=\underset{\frac{\pi }{4}}{\overset{{\sec }^{-1}\sqrt{e^x}}{\int }}\frac{2\tan \theta d\theta }{\sqrt{{\sec }^2\theta -1}}$
$\frac{\pi }{6}=\underset{\frac{\pi }{4}}{\overset{{\sec }^{-1}\sqrt{e^{x}2}}{\int }}\frac{2\tan \theta }{\tan \theta }d\theta =2{\left[\theta \right]}_{\pi /4}^{{\sec }^{-1\left(e^{x/2}\right)}}$
$\Rightarrow \frac{\pi }{6}=2\left[{\sec }^{-1\left(e^{x2}\right)}-\frac{\pi }{4}\right]$
$\Rightarrow {\sec }^{-1\left(e^{x2}\right)}-\frac{\pi }{4}$
$\Rightarrow e^{x/2}=\sec \frac{\pi }{3}=2$
Taking logarithms on both sides, we get
$\frac{x}{2}=\log 2\Rightarrow x=2\log 2=\log 2^2=\log 4$
$x=\log 4$