Question

The area of the region bounded by the curve $y=\frac{1}{1+{\left(\tan x\right)}^{\sqrt{2}}}$ and the $x-$axis between the ordinates $x=\frac{\pi }{6}$ and $x=\frac{\pi }{3}$ is $\frac{\pi }{12}$ sq units.

• A. True
• B. False

Answer 1

The correct option is A True

The given curve is $y=\frac{1}{1+{\left(\tan x\right)}^{\sqrt{2}}}$
$\Rightarrow y=\frac{{\left(\cos x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}$
Area bounded between the intervals $x=\frac{\pi }{6}$ and $x=\frac{\pi }{3}$ is given by
$A={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\left(\cos x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}dx$
Also $A=\frac{{\left(\sin x\right)}^{\sqrt{2}}}{{\left(\sin x\right)}^{\sqrt{2}}+{\left(\cos x\right)}^{\sqrt{2}}}$
Adding we get

$2A={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx$
$A=\frac{1}{2}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx$

$=\frac{1}{2}{\left[x\right]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}$

$=\frac{1}{2}[\frac{\pi }{3}-\frac{\pi }{6}]$

$=\frac{\pi }{12}$.sq.units.