The area of the region between the curves y=s | vedclass

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(B) The area $A$ is given by $\int_0^{\pi/4} \left( \sqrt{\frac{1+\sin x}{\cos x}} - \sqrt{\frac{1-\sin x}{\cos x}} \right) dx$. Using the half-angle

Vedclass · Accepted Answer

$\int_0^{\sqrt{2}-1} \frac{4t}{(1+t^2) \sqrt{1-t^2}} dt$

Vedclass · Answer

(B) The area $A$ is given by $\int_0^{\pi/4} \left( \sqrt{\frac{1+\sin x}{\cos x}} - \sqrt{\frac{1-\sin x}{\cos x}} ight) dx$. Using the half-angle identities $\sin x = \frac{2	an(x/2)}{1+	an^2(x/2)}$ and $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$,we simplify the integrand. Let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx = \frac{1}{2}(1+t^2) dx$,so $dx = \frac{2}{1+t^2} dt$. At $x=0$,$t=0$. At $x=\pi/4$,$t=	an(\pi/8) = \sqrt{2}-1$. The expression becomes $\int_0^{\sqrt{2}-1} \left( \sqrt{\frac{1+\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}} - \sqrt{\frac{1-\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}} ight) \frac{2}{1+t^2} dt$. This simplifies to $\int_0^{\sqrt{2}-1} \left( \sqrt{\frac{(1+t)^2}{1-t^2}} - \sqrt{\frac{(1-t)^2}{1-t^2}} ight) \frac{2}{1+t^2} dt = \int_0^{\sqrt{2}-1} \frac{2(1+t - (1-t))}{\sqrt{1-t^2}} \frac{1}{1+t^2} dt = \int_0^{\sqrt{2}-1} \frac{4t}{(1+t^2)\sqrt{1-t^2}} dt$.

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

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