Question

$x·\left(\frac{dy}{dx}\right)·\tan \left(\frac{y}{x}\right)=y\tan \left(\frac{y}{x}\right)+x$, $y\left(\frac{1}{2}\right)=\frac{\pi }{6}$. Area bounded by $x=0,x=\frac{1}{\sqrt{2}},y=y\left(x\right)$

Answer 1

Step 1: Note the given data

$y\left(\frac{1}{2}\right)=\frac{\pi }{6}$

The equation is $x·\left(\frac{dy}{dx}\right)·\tan \left(\frac{y}{x}\right)=y\tan \left(\frac{y}{x}\right)+x$

Dividing the equation by $x$

$\left(\frac{dy}{dx}\right)·\tan \left(\frac{y}{x}\right)=\frac{y}{x}\tan \left(\frac{y}{x}\right)+1........\left(1\right)$

Let $y=vx..........\left(2\right)$

Differentiating with respect to $x$

$\frac{dy}{dx}=v+x\frac{dv}{dx}............\left(3\right)$

Step 2: Substituting equations (2) and (3) in equation (1)

$\left(v+x\frac{dv}{dx}\right)·\tan v=v\tan v+1$

$\Rightarrow$$v\tan v+x\tan v\frac{dv}{dx}=v\tan v+1$

$\Rightarrow$ $\tan vdv=\frac{1}{x}dx$

Integrating both sides,

$\Rightarrow$ $\int \tan vdv=\int \frac{1}{x}dx$

$\Rightarrow$ $-\ln \left|\cos v\right|=\ln \left|x\right|+\ln c$

$\Rightarrow$ $-\cos v=xc$

$\Rightarrow$ $c=-\frac{\cos v}{x}$

$\Rightarrow$ $c=-\frac{\cos \left({\displaystyle \frac{y}{x}}\right)}{x}$

$\Rightarrow$ $\frac{y}{x}={\cos }^{-1}\left(-cx\right)$

$\Rightarrow$ '$\frac{y}{x}={\cos }^{-1}\left(cx\right)\left[\because {\cos }^{-1}\left(-x\right)={\cos }^{-1}\left(x\right)\right]$

$\Rightarrow$ $y=x{\cos }^{-1}\left(cx\right)$

Given that $y\left(\frac{1}{2}\right)=\frac{\pi }{6}$

$\Rightarrow$ $\frac{\mathrm{\pi }}{6}=\frac{1}{2}{\cos }^{-1}\left(\frac{1}{2}c\right)$

$\Rightarrow$ $\cos \frac{\mathrm{\pi }}{3}=\frac{1}{2}c$

$\Rightarrow$ $\frac{1}{2}=\frac{1}{2}c$

$\Rightarrow$ $c=1$

$\therefore y=x{\cos }^{-1}x$

Step 3: Find the area for the given boundary

Integration formula:

• $\int uvdx=v\int udx-\int \left\{\frac{dv}{dx}\int udx\right\}dx$
• $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$

Derivative formula

$\frac{d{\cos }^{-1}\left(x\right)}{dx}=-\frac{1}{\sqrt{1-x^2}}$

$A={\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}x{\cos }^{-1}xdx$

$\Rightarrow$ $A={\left[{\cos }^{-1}x·\frac{x^2}{2}\right]}_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}-{\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\frac{-1}{\sqrt{1-x^2}}\frac{x^2}{2}dx$

$\Rightarrow$ $A=\frac{1}{2}\left[\frac{\mathrm{\pi }}{4}·\frac{1}{2}\right]+\frac{1}{2}{\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\frac{x^2}{\sqrt{1-x^2}}dx$

$\Rightarrow$ $A=\frac{\mathrm{\pi }}{16}+\frac{1}{2}{\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\frac{x^2}{\sqrt{1-x^2}}dx$

Let $x=\cos \theta ;dx=-\sin \theta d\theta$

$x$	$\frac{1}{\sqrt{2}}$	$0$
$\theta$	$\frac{\pi }{4}$	$\frac{\pi }{2}$

$\Rightarrow$ $A=\frac{\mathrm{\pi }}{16}+\frac{1}{2}{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{-{\cos }^2\theta \sin \theta }{\sqrt{1-{\cos }^2\theta }}d\theta$

$\Rightarrow$ $A=\frac{\mathrm{\pi }}{16}-\frac{1}{2}{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{{\cos }^2\theta \sin \theta }{\sin \theta }d\theta$

$\Rightarrow$ $A=\frac{\mathrm{\pi }}{16}-\frac{1}{2}{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\cos }^2\theta d\theta$

$\Rightarrow$ $A=\frac{\mathrm{\pi }}{16}-\frac{1}{2}{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{\cos 2\theta +1}{2}d\theta \left[\because \cos 2\theta =2{\cos }^2\theta -1\right]$

$\Rightarrow$ $A=\frac{\mathrm{\pi }}{16}-\frac{1}{4}{\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left(\cos 2\theta +1\right)d\theta$

$\Rightarrow$$A=\frac{\mathrm{\pi }}{16}-\frac{1}{4}{\left[\frac{\sin 2\theta }{2}+\theta \right]}_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

$\Rightarrow$$A=\frac{\mathrm{\pi }}{16}-\frac{1}{4}\left[\left(\frac{1}{2}+\frac{\pi }{4}\right)-\left(0+\frac{\mathrm{\pi }}{2}\right)\right]$

$\Rightarrow$$A=\frac{\mathrm{\pi }}{16}-\frac{1}{4}\left(\frac{2+\mathrm{\pi }-2\mathrm{\pi }}{4}\right)$

$\Rightarrow$$A=\frac{\mathrm{\pi }}{16}-\frac{2-\mathrm{\pi }}{16}$

$\Rightarrow$$A=\frac{2\mathrm{\pi }-2}{16}$

$\Rightarrow$$A=\frac{\mathrm{\pi }-1}{8}sq.unit$

Hence the required area is $\frac{\mathrm{\pi }-1}{8}sq.unit$.