Question

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

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

Answer 1

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

$I=\underset{-\pi /4}{\overset{3\pi /4}{\int }}\frac{dx}{\sqrt{2}(e^{x-\pi /4}+1)\cos (x-\frac{\pi }{4})}$
Putting $x-\frac{\pi }{4}=t$, we get
$\Rightarrow I=\frac{1}{\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{dt}{(e^t+1)\cos t}\cdots \left(i\right)$
Using property $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)\Rightarrow I=\frac{1}{\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{e^{t}dt}{(e^t+1)\cos t}\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$2I=\frac{1}{\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\sec tdt\Rightarrow I=\frac{1}{2\sqrt{2}}\underset{-\pi /2}{\overset{\pi /2}{\int }}\sec xdx\therefore k=\frac{1}{2\sqrt{2}}$