Question

${\int }_{-\log 3}^{\log 3}{\cot }^{-1}\left(\frac{e^x-1}{e^x+1}\right)dx=$ _________.

• A. $\frac{\pi }{2}\log 3$
• B. $\pi \log 3$
• C. $0$
• D. $\pi \log 9$

Answer 1

Answer: (B)

$\pi \log 3$

$I={\int }_{-\log 3}^{\log 3}{\cot }^{-1}\left(\frac{e^x-1}{e^x+1}\right)dx$ ...... $\left(i\right)$

Let $f\left(x\right)={\cot }^{-1}\left(\frac{e^x-1}{e^x+1}\right)$

$\Rightarrow f(-x)={\cot }^{-1}\left(\frac{e^{-x}-1}{e^{-x}+1}\right)$

$\Rightarrow f(-x)={\cot }^{-1}\left(\frac{1-e^x}{1+e^x}\right)=f\left(x\right)$

$\therefore I={\int }_{-\log 3}^{\log 3}{\cot }^{-1}\left(\frac{1-e^x}{1+e^x}\right)dx$ ...... $\left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$2I=\pi {\int }_{-\log 3}^{\log 3}dx$ ........ $[\because {\cot }^{-1}x+{\cot }^{-1}y={\cot }^{-1}\left(\frac{xy-1}{x+y}\right)]$
$\Rightarrow I=\pi \log 3$.