Question

The area (in sq. units) bounded by curves $y=\text{cosec}x,$ $y=\sec x,$ $y=\cos x$ and $y=\sin x$ on interval $(0,\frac{\pi }{2})$ is equal to

• A. $1+\ln (\sqrt{2}+1)$
• B. $2\ln (\sqrt{2}+1)$
• C. $2\ln (\sqrt{2}+1)-\sqrt{2}$
• D. $\ln (\sqrt{2}+1)-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $2\ln (\sqrt{2}+1)-\sqrt{2}$


Area $=\underset{0}{\overset{\pi /4}{\int }}(\sec x-\cos x)dx+\underset{\pi /4}{\overset{\pi /2}{\int }}(\text{cosec}x-\sin x)dx$
$=\underset{0}{\overset{\pi /4}{\int }}\sec xdx-\underset{0}{\overset{\pi /4}{\int }}\cos xdx+\underset{\pi /4}{\overset{\pi /2}{\int }}\text{cosec}xdx-\underset{\pi /4}{\overset{\pi /2}{\int }}\sin xdx$
$=\ln (\sqrt{2}+1)-\frac{1}{\sqrt{2}}-\ln (\sqrt{2}-1)-\frac{1}{\sqrt{2}}$
$=2\ln (\sqrt{2}+1)-\sqrt{2}$