Mathematics - Area Under Curves Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Required area }={\int }_0^{\pi /4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right)dx\left[\because \frac{1+\sin x}{\cos x}>\frac{1-\sin x}{\cos x}>0\right]\\ & ={\int }_0^{\pi /4}\sqrt{\frac{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}}\sqrt{\frac{1-\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}})dx\\ & ={\int }_0^{\pi /4}\sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}})dx\\ & \end{array}$$ $$={\int }_0^{\pi /4}\frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{\sqrt{1-{\tan }^2\frac{x}{2}}}dx={\int }_0^{\pi /4}\frac{2\tan \frac{x}{2}}{\sqrt{1-{\tan }^2\frac{x}{2}}}dx$$ Put $\tan \frac{x}{2}=t$ $$\begin{array}{rl}\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx& =dt\\ \therefore \text{ Required area }& ={\int }_0^{\tan \frac{\pi }{8}}\frac{4tdt}{\left(1+t^2\right)\sqrt{1-t^2}}\\ & ={\int }_0^{\sqrt{2}-1}\frac{4t}{\left(1+t^2\right)\sqrt{1-t^2}}dt\left[\text{ astan }\frac{\pi }{8}=\sqrt{2}-1\right]\end{array}$$