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} $ and bounded by the
lines $x = 0$ and $ x = \frac{\pi}{4} $ is

• A. $ (a) \int \limits_0^{\sqrt 2 -1} \frac{t}{(1 + t^2) \sqrt {1 - t^2}} dt $
• B. $ (b) \int \limits_0^{\sqrt 2 -1} \frac{4t}{(1 + t^2) \sqrt {1 - t^2}} dt $
• C. $ (c) \int \limits_0^{\sqrt 2 +1} \frac{4t}{(1 + t^2) \sqrt {1 - t^2}} dt $
• D. $ (d) \int \limits_0^{\sqrt 2 +1} \frac{t}{(1 + t^2) \sqrt {1 - t^2}} dt $

Answer 1

Answer: (B)

Required area $ = \int \limits_0^{\pi/4} \bigg (\sqrt \frac{1 + sin x}{cos x} - \sqrt \frac{1 - sin x}{cos x}\bigg) dx$
$\, \, \, \, \, \, \, \, \, \, \, \, \Bigg [\because \frac{1 + sin x}{cos x} > \frac{1 - sin x}{cos x} > 0 \Bigg]$
$\, \, \, \, \, \, \, \, = \int \limits_0^{\pi/4} \Bigg ( \sqrt { \frac{ 1 + \frac{ 2 tan \, \frac{x}{2}}{ 1 + tan^2 \frac{x}{2}}}{ \frac{ 1 -tan^2 \frac{x}{2}}{ 1 + tan^2 \frac{x}{2}}}} - \sqrt { \frac{ 1 - \frac{ 2 tan \, \frac{x}{2}}{ 1 + tan^2 \frac{x}{2}}}{ \frac{ 1 -tan^2 \frac{x}{2}}{ 1 + tan^2 \frac{x}{2}}}} \Bigg) dx$
$ = \int \limits_0^{\pi/4} \Bigg (\sqrt \frac{1 + tan \frac{x}{2}}{1 - tan \frac{x}{2}} - \sqrt{ \frac{1 - tan \frac{x}{2}}{ 1 + tan \frac{x}{2}}} \Bigg) dx$
$ = \int \limits_0^{\pi/4} \frac{1 + tan\frac{x}{2} - 1 + tan\frac{x}{2}} {\sqrt {1 - tan^2\frac{x}{2}}} dx = \int \limits_0^{\pi/4} \frac{2tan\frac{x}{2}}{\sqrt{1 - tan^2\frac{x}{2}}} dx $
Put $ tan\frac{x}{2} = t \Rightarrow \frac{1}{2}sec^2\frac{x}{2} dx = dt $
$ \, \, \, \, \, \, = \int \limits_0^{tan{\pi/8}} \frac{4t dt}{(1 + t^2)\sqrt{1 - t^2}} $
As $\, \, \, \, \, \, \, \, \int \limits_0^{\sqrt2 - 1}\frac{4t dt}{(1 + t^2)\sqrt{1 - t^2}}\, \, \, [\because tan\frac{\pi}{8}= \sqrt 2 -1]$