Question

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

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

Answer 1

The correct option is B

${\int }_0^{\sqrt{2}-1}\frac{4t}{(1+t^2)\left(\sqrt{1-t^2}\right)}dt$

Explanation for the correct option:

Step 1: Finding the area of the given curve

Given curves are $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$
Boundary lines, $x=0,y=\frac{\pi }{4}$
We know in $x\in \left(0,\frac{\mathrm{\pi }}{4}\right)$ $\left(\sqrt{\frac{1+\sin x}{\cos x}}>\sqrt{\frac{1-\sin x}{\cos x}}\right)$
So, to find the area we need to integrate between zero and $\frac{\pi }{4}$.

Area $\begin{array}{rcl}& =& {\int }_0^{\pi /4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right)dx\end{array}$

Step 2: Evaluate $\sqrt{\frac{1-\sin x}{\cos x}}$:
$\begin{array}{rcl}& & \\ \frac{1-\sin x}{\cos x}& =& \frac{({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2})-\left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)}{({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})}\\ & =& \frac{{(\cos \frac{x}{2}-\sin \frac{x}{2})}^2}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}\\ & =& \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\\ & =& \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\end{array}$

Step 3: Evaluate $\sqrt{\frac{1+\sin x}{\cos x}}$:

Similarly, $\begin{array}{rcl}\frac{1+\sin x}{\cos x}& =& \frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}\end{array}$

$\begin{array}{rcl}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right)& =& {\int }_0^{\pi /4}\sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}}dx\\ & =& {\int }_0^{\pi /4}\left(\frac{\left(1+\tan \frac{x}{2}\right)-\left(1-\tan \frac{x}{2}\right)}{\sqrt{1-{\tan }^2\frac{x}{2}}}\right)dx\\ & =& {\int }_0^{\pi /4}\frac{2\tan \frac{x}{2}}{\sqrt{1-{\tan }^2\frac{x}{2}}}dx\end{array}$

Step 4: Evaluate $\begin{array}{rcl}& & {\int }_0^{\pi /4}\frac{2\tan \frac{x}{2}}{\sqrt{1-{\tan }^2\frac{x}{2}}}dx\end{array}$:

Let $\tan \frac{x}{2}=t$

$$\begin{gathered} {\sec }^2\frac{x}{2}\times \frac{1}{2}dx=dt \\ \left(1+{\tan }^2\frac{x}{2}\right)dx=2dt \\ \left(1+t^2\right)dx=2dt \\ dx=\frac{2dt}{1+t^2} \end{gathered}$$

The integral becomes,
$\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right)={\int }_0^{\sqrt{2}-1}\frac{4t}{\left(1+t^2\right)\sqrt[]{\left(1-t^2\right)}}dt$ $\left(\because \tan \frac{\pi }{8}=\sqrt{2}-1\right)$
Hence, option (B) is the correct answer.