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

(B) $\int_{0}^{\sqrt{2}-1} \frac{4 t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$

(A) $\int_{0}^{\sqrt{2}-1} \frac{t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$

(C) $\int_{0}^{\sqrt{2}+1} \frac{4 t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$

(D) $\int_{0}^{\sqrt{2}+1} \frac{t}{\left(1+t^{2}\right) \sqrt{1-t^{2}}} d t$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Area Under Curves

The area of the region between the curves

y = \frac{1 + s i n x}{c o s x}

and

y = \frac{1 - s i n x}{c o s x}

bounded by the lines

x = 0

and

x = \frac{π}{4}

Answer: (B)

The area is

\int_{0}^{π /4} (\frac{1 + s i n x}{c o s x} - \frac{1 - s i n x}{c o s x}) d x

. Using half-angle substitution

tan (x /2) = t

, we have

sin x = \frac{2 t}{1 + t ^{2}}

cos x = \frac{1 - t ^{2}}{1 + t ^{2}}

, and

d x = \frac{2 d t}{1 + t ^{2}}

. The expression simplifies to

\int_{0}^{2 - 1} \frac{2 t a n ( x /2 )}{1 - t a n ^{2} ( x /2 )} d x = \int_{0}^{2 - 1} \frac{4 t}{( 1 + t ^{2} ) 1 - t ^{2}} d t

. When

x = π /4

t = tan (π /8) = 2 - 1

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option