Question

Find the area of the region bounded by the $y-$axis, $y=\cos x$ and $y=\sin x,0\le x\le \pi /2$

Answer 1

We have,

$y=\cos xandy=\sin x$

Now,

$\sin x=\cos x$

$\frac{\sin x}{\cos x}=1$

$\Rightarrow \tan x=1$

$\Rightarrow \tan x=\tan \frac{\pi }{4}$

$\Rightarrow x=\frac{\pi }{4}$

Then,

At $x=\frac{\pi }{4}$, both are equal

So,

$y=\cos x=\cos \frac{\pi }{4}$

$y=\frac{1}{\sqrt{2}}$

So, Finding area

$ \text{Requried}\,\text{area = Area of ABCO-Area of BCO} $

$\text{Requried}\text{area =}{\int }_0^{\frac{\pi }{4}}ydx-{\int }_0^{\frac{\pi }{4}}ydx$

$AreaABCO={\int }_0^{\frac{\pi }{2}}\cos xdx$

$={{\left[\sin x\right]}_0}^{\frac{\pi }{4}}$

$=[\sin \frac{\pi }{4}-\sin 0]$

$=\frac{1}{\sqrt{2}}-0$

$=\frac{1}{\sqrt{2}}$

Area $BCO$ $={\int }_0^{\frac{\pi }{4}}ydx$

$={\int }_0^{\frac{\pi }{4}}\sin xdx$

$={{[-\cos x]}_0}^{\frac{\pi }{4}}$

$=-[\cos \frac{\pi }{4}-\cos 0]$

$=-[\frac{1}{\sqrt{2}}-1]$

$=1-\frac{1}{\sqrt{2}}$

Now,

Required area

$=\frac{1}{\sqrt{2}}-[1-\frac{1}{\sqrt{2}}]$

$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1$

$=\frac{2}{\sqrt{2}}-1$

$=\sqrt{2}-1$

Hence, this is the answer.