Area bounded by $y=\sqrt{\frac{1+\sin{x}}{\cos{x}}}$ and $y=\sqrt{\frac{1-\sin{x}}{\cos{x}}}$

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

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

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

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

My Attempt:

So I started with,

$$\int_{0}^{\frac{\pi}{4}}\sqrt{\frac{1+\sin{x}}{\cos{x}}}-\sqrt{\frac{1-\sin{x}}{\cos{x}}}dx$$

$$\int_{0}^{\frac{\pi}{4}}\frac{2\sin{x}}{\sqrt{\cos{x}}(\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}})}dx$$

Putting $\cos{x}=u$

$$\int_{1}^{\frac{1}{\sqrt{2}}}\frac{-2}{\sqrt{u}(\sqrt{1+\sqrt{1-u^2}}+\sqrt{1-\sqrt{1-u^2}})}du$$

Putting $u=\frac{1}{w}$ and further $w-1=t$,

I end up with,

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

But I am not able to reduce this to any of the given options? Am I miscalculating something or this is reducible?

Any hints would be helpful.

Thank you.

Answer 1

HINT

We could use Tangent half-angle substitution to obtain

$$\sin x=\frac{2t}{1+t^2}\quad \cos x=\frac{1-t^2}{1+t^2}\quad t=\tan \frac x 2 \implies dt=\frac12(1+t^2)dx$$

then

$$\int_{0}^{\frac{\pi}{4}}\sqrt{\frac{1+\sin{x}}{\cos{x}}}-\sqrt{\frac{1-\sin{x}}{\cos{x}}}dx=\int_0^{\sqrt2 -1} \left(\sqrt{\frac{(1+t)^2}{1-t^2}}-\sqrt{\frac{(1-t)^2}{1-t^2}}\right)\frac2{1+t^2}dt=\ldots$$

Answer 2

$$I=\int_0^{\pi/4}\sqrt{\frac{1+\sin(x)}{\cos(x)}}-\sqrt{\frac{1-\sin(x)}{\cos(x)}}dx$$ using Weierstrass substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution

we know that: $$\sin(x)=\frac{2t}{1+t^2},\,\cos(x)=\frac{1-t^2}{1+t^2},\,dx=\frac{2dt}{1+t^2}$$ so we can obtain: $$I=\int_0^{\sqrt{2}-1}\left(\sqrt{\frac{1+t^2+2t}{1-t^2}}-\sqrt{\frac{1+t^2-2t}{1-t^2}}\right).\frac{2dt}{1+t^2}$$$$ =\int_0^{\sqrt{2}-1}\left(\sqrt{\frac{(1+t)^2}{(1-t)(1+t)}}-\sqrt{\frac{(1-t)^2}{(1-t)(1+t)}}\right).\frac{2dt}{1+t^2}$$$$ =\int_0^{\sqrt{2}-1}\left(\sqrt{\frac{1+t}{1-t}}-\sqrt{\frac{1-t}{1+t}}\right).\frac{2dt}{1+t^2}$$$$ =\int_0^{\sqrt{2}-1}\frac{(1+t)-(1-t)}{\sqrt{1-t^2}}.\frac{2dt}{1+t^2}$$$$ =\int_0^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$$ so it is $(b)$