IIT-JEE 2008 · Paper 2, Mathematics question 4

Half-angle substitution. With $t=\tan\frac x2$: $\dfrac{1\pm\sin x}{\cos x}=\dfrac{(1\pm t)^2}{1-t^2}=\dfrac{1\pm t}{1\mp t}$.

Difference of the curves (the first is above on $[0,\frac\pi4]$):
$$\sqrt{\frac{1+t}{1-t}}-\sqrt{\frac{1-t}{1+t}}=\frac{2t}{\sqrt{1-t^2}}.$$

Area. $dx=\dfrac{2\,dt}{1+t^2}$, and $x=\frac\pi4$ gives $t=\tan\frac\pi8=\sqrt2-1$:
$$\int_0^{\sqrt2-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt.$$

(B)