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. $\int\limits_0^{\sqrt{2}-1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$
• B. $\int\limits_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$
• C. $\int\limits_0^{\sqrt{2}+1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$
• D. $\int\limits_0^{\sqrt{2}+1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

Answer 1

Answer: (B)

Area $=\int\limits_0^{\pi / 4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right) d x $
$ =\int\limits_0^{\pi / 4}\left(\sqrt{\frac{1+\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}}{\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)}}}-\sqrt{\frac{1-\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}}{\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)}}}\right) d x $
$=\int\limits_0^{\pi / 4} \frac{2 \tan (x / 2)}{\sqrt{1-\tan ^2(x / 2)}} d x $
Let $\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t$
by substitution Integration becomes
$\int\limits_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$