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 $x=\frac{π}{4}$ is

• A. $\underset{0}{\overset{\sqrt{2}−1}{∫}}\frac{t}{\left(1+t^2\right)\sqrt{1−t^2}}dt$
• B. $\underset{0}{\overset{\sqrt{2}−1}{∫}}\frac{4t}{\left(1+t^2\right)\sqrt{1−t^2}}dt$
• C. $\underset{0}{\overset{\sqrt{2}+1}{∫}}\frac{4t}{\left(1+t^2\right)\sqrt{1−t^2}}dt$
• D. $\underset{0}{\overset{\sqrt{2}+1}{∫}}\frac{t}{\left(1+t^2\right)\sqrt{1−t^2}}dt$

Answer 1

Answer: (B)

Area $=\underset{0}{\overset{π/4}{∫}}\left(\sqrt{\frac{1+\sin x}{\cos x}}−\sqrt{\frac{1−\sin x}{\cos x}}\right)dx$
$=\underset{0}{\overset{π/4}{∫}}\left(\sqrt{\frac{1+\frac{2\tan (x/2)}{1+{\tan }^2(x/2)}}{\frac{1−{\tan }^2(x/2)}{1+{\tan }^2(x/2)}}}−\sqrt{\frac{1−\frac{2\tan (x/2)}{1+{\tan }^2(x/2)}}{\frac{1−{\tan }^2(x/2)}{1+{\tan }^2(x/2)}}}\right)dx$
$=\underset{0}{\overset{π/4}{∫}}\frac{2\tan (x/2)}{\sqrt{1−{\tan }^2(x/2)}}dx$
Let $\tan \frac{x}{2}=t⇒\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
by substitution Integration becomes
$\underset{0}{\overset{\sqrt{2}−1}{∫}}\frac{4t}{\left(1+t^2\right)\sqrt{1−t^2}}dt$