Question

Value of ${\int }_{\pi /2}^{5\pi /2}\frac{e^{ta{n}^{-1}\left(sinx\right)}}{e^{ta{n}^{-1}\left(sinx\right)}+e^{ta{n}^{-1}\left(cosx\right)}}$ dx is equal to:

• A. $3\pi$
• B. $2\pi$
• C. $\pi$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (C)

$\pi$

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

$I_1={\int }_0^{5\pi /2}\frac{e^{te{n}^{-1}\left(\sin x\right)}}{e^{{\tan }^{-1}\left(sinx\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx------\left(i\right)$

Using formula$\Rightarrow \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f\left(-x\right)dx$

$I_1=\underset{0}{\overset{5\pi /2}{\int }}\frac{e^{{\tan }^{-1}\left(\cos x\right)}}{e^{{\tan }^{-1}\left(\cos x\right)}+e^{te{n}^{-1}\left(\sin x\right)}}dx-----\left(ii\right)$

Using $\cos \left(5\pi /2-x\right)=\sin x,\sin \left(5\pi /2-x\right)=\cos x$

$Adding\left(i\right)or\left(ii\right)$

$2I_1={\int }_0^{5\pi /2}\frac{e^{{\tan }^{-1}\left(\sin x\right)}+e^{te{n}^{-1}\left(\cos x\right)}}{e^{te{n}^{-1}\left(\cos x\right)}+e^{{\tan }^{-1}\left(\sin x\right)}}dx$

$\begin{array}{c}\Rightarrow 2I_1=\frac{5\pi }{2}-0\\ \Rightarrow I_1=\frac{5\pi }{4}\end{array}$

$I_2={\int }_0^{\pi /2}\frac{e^{te{n}^{-1}\left(\sin x\right)}}{e^{te{n}^{-1}\left(\cos x\right)}+e^{te{n}^{-1}\left(\sin x\right)}}dx----\left(iii\right)$

Again using formula ${\int }_0^{a}f\left(n\right)dx={\int }_0^{a}f\left(a-x\right)dx$

$I_2={\int }_0^{\pi /2}\frac{e^{te{n}^{-1}\left(\cos x\right)}}{e^{te{n}^{-1}\left(\sin x\right)}+e^{te{n}^{-1}\left(\cos x\right)}}-----\left(iv\right)$

$\because \sin \left(\pi /2-x\right)=\cos x$

$Adding\left(iii\right)or\left(iv\right).$

$2I_2={\int }_0^{\pi /2}\frac{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}{e^{{\tan }^{-1}\left(\sin x\right)}+e^{{\tan }^{-1}\left(\cos x\right)}}dx$

$\Rightarrow I_2=\pi /4$

So,

$I={\int }_{\pi /2}^0+{\int }_0^{5\pi /2}=-{\int }_0^{\pi /2}+{\int }_0^{5\pi 2}$

$\begin{array}{c}\Rightarrow I=-\pi /4+5\pi 4\\ \Rightarrow I=\frac{4\pi }{4}\\ \Rightarrow I=\pi \end{array}$

Hence,

$C$ is correct options