Question

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

• A. $1$
• B. $\pi$
• C. $e$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\pi$

$I=\underset{\pi /2}{\overset{5\pi /2}{\int }}\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx\Rightarrow I=\underset{\pi /2}{\overset{2\pi }{\int }}\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx+\underset{2\pi }{\overset{5\pi /2}{\int }}\frac{e^{{\tan }^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx\cdots \left(1\right)\text{Using}\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx\Rightarrow I=\underset{\pi /2}{\overset{2\pi }{\int }}\frac{e^{{\tan }^{-1}\left(\sin \right(2\pi +\frac{\pi }{2}-x\left)\right)}}{e^{{\tan }^{-1}\left(\sin \right(2\pi +\frac{\pi }{2}-x\left)\right)}+e^{{\tan }^{-1}\left(\cos \right(2\pi +\frac{\pi }{2}-x\left)\right)}}dx+\underset{2\pi }{\overset{5\pi /2}{\int }}\frac{e^{{\tan }^{-1}\left(\sin \right(2\pi +\frac{5\pi }{2}-x\left)\right)}}{e^{{\tan }^{-1}\left(\sin \right(2\pi +\frac{5\pi }{2}-x\left)\right)}+e^{{\tan }^{-1}\left(\cos \right(2\pi +\frac{\pi }{2}-x\left)\right)}}dx\Rightarrow I=\underset{\pi /2}{\overset{2\pi }{\int }}\frac{e^{{\tan }^{-1}\left(\cos x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx+\underset{\pi }{\overset{5\pi /2}{\int }}\frac{e^{{\tan }^{-1}\left(\cos x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx\cdots \left(2\right)$
$\left(1\right)+\left(2\right)$
$\Rightarrow 2I=\underset{\pi /2}{\overset{2\pi }{\int }}1dx+\underset{2\pi }{\overset{5\pi /2}{\int }}1dx\Rightarrow 2I=x{|}_{\pi /2}^{5\pi /2}=2\pi \Rightarrow I=\pi$