Question

The value of $I={\int }_{\pi /2}^{5\pi /2}\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx$ is

• A. $1$
• B. $\pi$
• C. $e$
• D. $\pi /2$

Answer 1

Answer: (B)

$\pi$

$I={\int }_{\pi /2}^{5\pi /2}\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx$

$I={\int }_{\pi /2}^{\pi }\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx+{\int }_{\pi }^{5\pi /2}\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx.....\left(1\right)$

$I={\int }_{\pi /2}^{\pi }\frac{e^{-{\tan }^{-1}\left(\sin x\right)}}{e^{-{\tan }^{-1}\left(\sin x\right)}+e^{-{\tan }^{-1}\left(\cos x\right)}}dx+{\int }_{\pi }^{5\pi /2}\frac{e^{-{\tan }^{-1}\left(\sin x\right)}}{e^{-{\tan }^{-1}\left(\sin x\right)}+e^{-{\tan }^{-1}\left(\cos x\right)}}dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]$

$I={\int }_{\pi /2}^{\pi }\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx+{\int }_{\pi }^{5\pi /2}\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx......\left(2\right)$

from (1)+(2)

$2I={\int }_{\pi /2}^{\pi }1dx+{\int }_{\pi }^{5\pi /2}1dx2I={\left(x\right)}_{\pi /2}^{5\pi /2}=2\pi$

$\Rightarrow I=\pi$