Area bounded by two curve and the line $x=0$ and $x=\frac{\pi}{4}$

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}} }}dt} $

(B) $\int\limits_0^{\sqrt 2 - 1} {\frac{{4t}}{{\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} $

(C) $\int\limits_0^{\sqrt 2 + 1} {\frac{{4t}}{{\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} $

(D)$\int\limits_0^{\sqrt 2 + 1} {\frac{t}{{\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} $

The official answer is (B).

How do i proceed with this question

Answer 1

You don't need to sketch the graph .

$f(x) = \sqrt {\frac{{1 + \sin x}}{{\cos x}}} $ and $g(x) = \sqrt {\frac{{1 - \sin x}}{{\cos x}}} $

observe that for $0\le x\le \pi/4$ $$f^2(x)-g^2(x)=2\tan x\ge 0 $$ $$\Rightarrow f(x)\ge g(x)$$

This means $f(x)$ always lies above the graph of $g(x)$ for given interval.

we require $$\int_{0}^{\pi/4} (f(x)-g(x))dx=?$$.

I leave this to you to figure out the apt substituition