Question

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

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

Answer 1

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

$I={\int }_{-\pi /4}^{3\pi /4}\frac{dx}{\sqrt{2}(e^{(x-\frac{\pi }{4})}+1)\cos (x-\frac{\pi }{4})}$
Putting $x-\frac{\pi }{4}=t$, we get
$I=\frac{1}{\sqrt{2}}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dt}{(e^t+1)\cos t}$
$=\frac{1}{\sqrt{2}}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{e^{t}dt}{(e^t+1)\cos t}$
Adding, we get $2I=\frac{1}{\sqrt{2}}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sec tdt$
$\therefore I=\frac{1}{2\sqrt{2}}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sec xdx$ $\therefore k=\frac{1}{2\sqrt{2}}$