Question

If $\underset{-\pi /4}{\overset{3\pi /4}{\int }}\frac{e^{\pi /4}dx}{(e^x+e^{\pi /4})(\sin x+\cos x)}=\lambda \underset{-\pi /2}{\overset{\pi /2}{\int }}\sec xdx,$ then $\lambda$ is equal to

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

Answer 1

The correct option is C $\frac{1}{2\sqrt{2}}$

$I=\underset{-\pi /4}{\overset{3\pi /4}{\int }}\frac{e^{\pi /4}dx}{(e^x+e^{\pi /4})(\sin x+\cos x)}=\lambda \underset{-\pi /2}{\overset{\pi /2}{\int }}\sec xdx$

$\Rightarrow I=\underset{-\pi /4}{\overset{3\pi /4}{\int }}\frac{e^{\pi /4}dx}{e^x(1+e^{\pi /4-x})(\sin x+\cos x)}$

$\Rightarrow I=\underset{-\pi /4}{\overset{3\pi /4}{\int }}\frac{e^{\pi /4-x}dx}{(1+e^{\pi /4-x})(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)\sqrt{2}}$

$\Rightarrow I=\frac{1}{\sqrt{2}}\underset{-\pi /4}{\overset{3\pi /4}{\int }}\frac{e^{\pi /4-x}dx}{(1+e^{\pi /4-x})\left[\cos (\frac{\pi }{4}-x)\right]}$

Let $\frac{\pi }{4}-x=t$
$\Rightarrow dx=-dt$

$I=-\frac{1}{\sqrt{2}}\underset{\pi /2}{\overset{-\pi /2}{\int }}\frac{e^t}{(1+e^t)\cos t}dt$

$\Rightarrow I=\frac{1}{\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{e^t}{(1+e^t)\cos t}dt\cdots \left(1\right)$

Applying king's property,
$I=\frac{1}{\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{e^{-t}}{(1+e^{-t})\cos (-t)}dt$

$\Rightarrow I=\frac{1}{\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{1}{(1+e^t)\cos t}dt\cdots \left(2\right)$

On adding equations $\left(1\right)$ and $\left(2\right),$
$2I=\frac{1}{\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\sec tdt$

$\Rightarrow I=\frac{1}{2\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\sec xdx=\lambda \underset{-\pi /2}{\overset{\pi /2}{\int }}\sec xdx$

On comparing, $\lambda =\frac{1}{2\sqrt{2}}$