Question

The area of the region between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac{\pi }{4}$ is

• A. ${\int }_0^{\sqrt{2}-1}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$
• B. ${\int }_0^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
• C. ${\int }_0^{\sqrt{2}+1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
• D. ${\int }_0^{\sqrt{2}+1}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$

Answer 1

The correct option is B ${\int }_0^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$

${\int }_0^{\pi /4}(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}})dx$
$={\int }_0^{\pi /4}(\sqrt{\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}}-\sqrt{\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}})dx=\int \frac{(1+tan\frac{x}{2})-(1-tan\frac{x}{2})}{\sqrt{1-ta{n}^2\frac{x}{2}}}dx$
$={\int }_0^{\pi /4}\frac{2\tan \frac{x}{2}}{\sqrt{1-{\tan }^2\frac{x}{2}}}dx={\int }_0^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$ as $\tan \frac{x}{2}=t$.