Nice Exercise about the range using Trigonometric substitution

Exercise:

Let $f$ be a function defined to be $f\left( x \right)=\frac{1+\sqrt{\frac{1}{x}}}{\sqrt{1+\frac{1}{x}}}$

1.       Determine the range and the Domain of $f$
2.    Compute, $\int_{0}^{1}{f\left( x \right)dx}$

Solution:  

(1)    we have $f\left( x \right)=\frac{1+\sqrt{\frac{1}{x}}}{\sqrt{1+\frac{1}{x}}}$ is defined when $1+\frac{1}{x}>0$

So $1+\frac{1}{x}=\frac{x+1}{x}>0$ but $x\ne 0$ so $x>0\,\,or\,\,x<-1$

But $\sqrt{\frac{1}{x}}$ is defined when $x>0$ thus $dom\left( f \right)=\left\{ x\in \mathbb{R}:x>0 \right\}$

Let $x={{\tan }^{2}}\theta $ and we know that ${{\tan }^{2}}\theta >0$ and ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1>0$

Hence $f\left( x \right)=f\left( {{\tan }^{2}}\theta  \right)=\frac{1+\sqrt{\frac{1}{{{\tan }^{2}}\theta }}}{\sqrt{1+\frac{1}{{{\tan }^{2}}\theta }}}=\frac{1+\frac{1}{\tan \theta }}{\frac{\sqrt{{{\tan }^{2}}\theta +1}}{\tan \theta }}=\frac{\tan \theta +1}{\sec \theta }=\frac{\tan \theta }{\sec \theta }+\frac{1}{\sec \theta }$

$=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}+\frac{1}{\frac{1}{\cos \theta }}=\sin \theta +\cos \theta $ thus $f\left( {{\tan }^{2}}\theta  \right)=\sin \theta +\cos \theta $

But $\sin \theta +\cos \theta =\frac{\sqrt{2}}{\sqrt{2}}\sin \theta +\frac{\sqrt{2}}{\sqrt{2}}\cos \theta =\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin \theta +\cos \theta \frac{1}{\sqrt{2}} \right)=\sqrt{2}\sin \left( \theta +\frac{\pi }{4} \right)$

But $-1\le \sin u\left( \theta  \right)\le 1$ where $u\left( \theta  \right)=\theta +\frac{\pi }{4}$ $\Leftrightarrow -\sqrt{2}\le \sqrt{2}\sin u\left( \theta  \right)\le \sqrt{2}$

Thus $-\sqrt{2}\le f\left( {{\tan }^{2}}\theta  \right)\le \sqrt{2}$ but ${{\tan }^{2}}\theta >0\,\,\,\And \,\,{{\sec }^{2}}\theta >1\Leftrightarrow 1\le f\left( x \right)\le \sqrt{2}$

Therefore $Range\left( f \right)=\left\{ y\in \mathbb{R}:y\le \sqrt{2}\,\,\And \,y\ge 1 \right\}$

(2) $\int_{0}^{1}{f\left( x \right)dx}=??$

We have $1+\sqrt{\frac{1}{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}$ and $\sqrt{1+\frac{1}{x}}=\sqrt{\frac{x+1}{x}}=\frac{\sqrt{x+1}}{\sqrt{x}}$

So $\frac{1+\sqrt{\frac{1}{x}}}{\sqrt{1+\frac{1}{x}}}=\frac{\frac{\sqrt{x}+1}{\sqrt{x}}}{\frac{\sqrt{x+1}}{\sqrt{x}}}=\frac{\sqrt{x}+1}{\sqrt{x+1}}$

So $\int_{0}^{1}{f\left( x \right)dx}=\int_{0}^{1}{\frac{\sqrt{x}+1}{\sqrt{x+1}}dx}=\int_{0}^{1}{\frac{\sqrt{x}}{\sqrt{x+1}}dx}+\int_{0}^{1}{\frac{dx}{\sqrt{x+1}}}$

Let $u=x+1\Rightarrow du=dx$ so \(\int_{0}^{1}{\frac{dx}{\sqrt{x+1}}=\int_{1}^{2}{\frac{du}{\sqrt{u}}=\left[ 2\sqrt{u} \right]_{1}^{2}=2\sqrt{2}-2}}\)

We have $\frac{\sqrt{x}}{\sqrt{x+1}}=\frac{\sqrt{x}}{\sqrt{x\left( 1+\frac{1}{x} \right)}}=\frac{\sqrt{x}}{\sqrt{x}\sqrt{1+\frac{1}{x}}}=\frac{1}{\sqrt{1+\frac{1}{x}}}$ So $\int_{0}^{1}{\frac{\sqrt{x}}{\sqrt{x+1}}dx}=\int_{0}^{1}{\frac{dx}{\sqrt{1+\frac{1}{x}}}}$

Let $u=\frac{1}{x}\Rightarrow du=-\frac{1}{{{x}^{2}}}dx\Rightarrow -{{x}^{2}}du=dx$

So $\int{\frac{dx}{\sqrt{1+\frac{1}{x}}}=\int{\frac{-{{x}^{2}}du}{\sqrt{1+u}}}}=-\int{\frac{\frac{1}{{{u}^{2}}}}{\sqrt{1+u}}du}=-\int{\frac{du}{{{u}^{2}}\sqrt{1+u}}}$

Let $v=\sqrt{1+u}\Rightarrow dv=\frac{1}{2\sqrt{1+u}}du\Rightarrow 2vdv=du$

So $-\int{\frac{du}{{{u}^{2}}\sqrt{1+u}}=-\int{\frac{2vdv}{{{\left( {{v}^{2}}-1 \right)}^{2}}v}=-\int{\frac{2dv}{{{\left( {{v}^{2}}-1 \right)}^{2}}}}}}=-\int{\frac{2}{{{\left( v-1 \right)}^{2}}{{\left( v+1 \right)}^{2}}}dv}$

But $\frac{1}{{{\left( v+1 \right)}^{2}}{{\left( v-1 \right)}^{2}}}=\frac{1}{4\left( v+1 \right)}+\frac{1}{4{{\left( v+1 \right)}^{2}}}-\frac{1}{4\left( v-1 \right)}+\frac{1}{4{{\left( v-1 \right)}^{2}}}$

So $-\int{\frac{2}{{{\left( {{v}^{2}}-1 \right)}^{2}}}dv}=-\frac{2}{4}\ln \left| v+1 \right|-\frac{2}{4}\int{\frac{dv}{{{\left( v+1 \right)}^{2}}}+\frac{2}{4}\ln \left| v-1 \right|-\frac{2}{4}\int{\frac{dv}{{{\left( v-1 \right)}^{2}}}}}$

                          \(=-\frac{1}{2}\ln \left( v+1 \right)+\frac{1}{2\left( v+1 \right)}+\frac{1}{2}\ln \left( v-1 \right)+\frac{1}{2\left( v-1 \right)}+c\)

                          $=-\frac{1}{2}\ln \left( \sqrt{1+\frac{1}{x}}+1 \right)+\frac{1}{2\left( \sqrt{1+\frac{1}{x}}+1 \right)}+\frac{1}{2}\ln \left( \sqrt{1+\frac{1}{x}}-1 \right)+\frac{1}{2\left( \sqrt{1+\frac{1}{x}}-1 \right)}+c$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( -\frac{1}{2}\ln \left( \sqrt{1+\frac{1}{x}}+1 \right)+\frac{1}{2}\ln \left( \sqrt{1+\frac{1}{x}}-1 \right) \right)=-\frac{1}{2}\ln \left( \sqrt{2}+1 \right)+\frac{1}{2}\ln \left( \sqrt{2}-1 \right)$

But $\ln \left( \sqrt{2}+1 \right)=\operatorname{arcsinh}\left( 1 \right)\,\And \,\ln \left( \sqrt{2}-1 \right)=-\operatorname{arcsinh}\left( 1 \right)$

Thus $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( \frac{-1}{2}\ln \left( \sqrt{1+\frac{1}{x}}+1 \right)+\frac{1}{2}\ln \left( \sqrt{1+\frac{1}{x}}-1 \right) \right)=-\frac{\operatorname{arcsinh}\left( 1 \right)}{2}-\frac{\operatorname{arcsinh}\left( 1 \right)}{2}=-\operatorname{arcsinh}\left( 1 \right)$

And $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\left( \frac{1}{2\left( \sqrt{1+\frac{1}{x}}+1 \right)}+\frac{1}{2\left( \sqrt{1+\frac{1}{x}}-1 \right)} \right)=\frac{1}{2\left( \sqrt{2}+1 \right)}+\frac{1}{2\left( \sqrt{2}-1 \right)}=\frac{\sqrt{2}-1+\sqrt{2}+1}{2\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}$

$=\frac{2\sqrt{2}}{2\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}=\frac{\sqrt{2}}{2-1}=\sqrt{2}$ So $\int_{0}^{1}{\frac{\sqrt{x}}{\sqrt{x+1}}dx}=\sqrt{2}-\operatorname{arcsinh}\left( 1 \right)$

Therefore $\int_{0}^{1}{f\left( x \right)dx}=\sqrt{2}-\operatorname{arcsinh}\left( 1 \right)+2\sqrt{2}-2=3\sqrt{2}-2-\operatorname{arcsinh}\left( 1 \right)$ 

 *The idea of solution in finding the range is Credit to فراس الناصري‎*