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 (1) $\int_{0}^{\sqrt{2}-1}\frac{t}{(1+t^{2})\sqrt{1-t^{2}}}\,dt$ (2) $\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^{2})\sqrt{1-t^{2}}}\,dt$ (3) $\int_{0}^{\sqrt{2}+1}\frac{4t}{(1+t^{2})\sqrt{1-t^{2}}}\,dt$ (4) $\int_{0}^{\sqrt{2}+1}\frac{t}{(1+t^{2})\sqrt{1-t^{2}}}\,dt$

The area of the region between the curves y=√(1+sin x)/(cos x) and y=√(1-sin x)/(cos x) bounded by the lines x=0 and x=π/4 is $y_1 = \sqrt{\dfrac{1+\sin x}{\cos x}}, \qquad y_2 = \sqrt{\dfrac{1-\sin x}{\cos x}}$ $x = 0, \qquad x = \dfrac{\pi}{4}$ $\text{To find: the area } A \text{ between the curves on } \left[0, \tfrac{\pi}{4}\right]$ $A = \displaystyle\int_{0}^{\pi/4} (y_1 - y_2)\,dx$ $\text{use the half-angle substitution } t = \tan\tfrac{x}{2}$ $y_1 y_2 = \sqrt{\dfrac{(1+\sin x)(1-\sin x)}{\cos^{2}x}} = \sqrt{\dfrac{\cos^{2}x}{\cos^{2}x}} = 1$ $y_1 = \sqrt{\tan\!\left(\tfrac{\pi}{4}+\tfrac{x}{2}\right)}, \quad y_2 = \sqrt{\tan\!\left(\tfrac{\pi}{4}-\tfrac{x}{2}\right)}$ $t = \tan\dfrac{x}{2} \Rightarrow x = 2\arctan t, \qquad dx = \dfrac{2}{1+t^{2}}\,dt$ $\sin x = \dfrac{2t}{1+t^{2}}, \qquad \cos x = \dfrac{1-t^{2}}{1+t^{2}}$ $y_1^{2} = \dfrac{1+\sin x}{\cos x} = \dfrac{(1+t)^{2}}{1-t^{2}} = \dfrac{1+t}{1-t}$ $y_2^{2} = \dfrac{1-t}{1+t} \Rightarrow y_1 = \sqrt{\dfrac{1+t}{1-t}}, \quad y_2 = \sqrt{\dfrac{1-t}{1+t}}$ $y_1 - y_2 = \dfrac{(1+t)-(1-t)}{\sqrt{1-t^{2}}} = \dfrac{2t}{\sqrt{1-t^{2}}}$ $x = 0 \Rightarrow t = 0; \qquad x = \dfrac{\pi}{4} \Rightarrow t = \tan\dfrac{\pi}{8} = \sqrt{2}-1$ $A = \displaystyle\int_{0}^{\pi/4}(y_1-y_2)\,dx = \int_{0}^{\sqrt{2}-1} \dfrac{2t}{\sqrt{1-t^{2}}}\cdot\dfrac{2}{1+t^{2}}\,dt$ $= \displaystyle\int_{0}^{\sqrt{2}-1} \dfrac{4t}{(1+t^{2})\sqrt{1-t^{2}}}\,dt$ $\Rightarrow \text{match: option (2)}$ $\boxed{\text{Answer: (2)}}$