Question

The value of ${\int }_0^{x/4}\frac{\sec x}{{(\sec x+\tan x)}^2}dx$ is

• A. $1+\sqrt{2}$
• B. $-11+\sqrt{2}$
• C. $-\sqrt{2}$
• D. None of these

Answer 1

The correct option is D None of these

$I={\int }_0^{\frac{\pi }{4}}\frac{\sec x}{{(\sec x+\tan x)}^2}dx$

$={\int }_0^{\frac{\pi }{4}}\frac{1/\cos x}{{[\frac{1}{\cos x}+\frac{\sin x}{\cos x}]}^2}dx$

$={\int }_0^{\frac{\pi }{4}}\frac{{\cos }^{2}x/\cos x}{{(1+\sin x)}^2}dx$
$I={\int }_0^{\frac{\pi }{4}}{(1+\sin x)}^{-2}\cos xdx$

Take $1+\sin x=t⟹\cos xdx=dt$

When $x=0,t=1$ and $x=\frac{\pi }{4},t=1+\frac{1}{\sqrt{2}}$
$\therefore I={\int }_1^{1+\frac{1}{\sqrt{2}}}{t}^{-2}dt$

$\therefore I={\left|\frac{t^{-1}}{-1}\right|}_1^{1+\frac{1}{\sqrt{2}}}$
$=-[\frac{1}{1+\frac{1}{\sqrt{2}}}-1]$
$=-\frac{\sqrt{2}}{\sqrt{2}+1}+1$$=\frac{-\sqrt{2}+\sqrt{2}+1}{\sqrt{2}+1}=\frac{1}{\sqrt{2}+1}$
$=\frac{1}{\sqrt{2}+1}\ast \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}-1}{2-1}$
$=\sqrt{2}-1$
$\therefore$ None of these is correct option