Question

Integrate: $\sqrt{\dfrac{2-x}{x}}$ $\left(0< x<2\right)$.

Answer 1

Hint: For this problem we will first take the substitution as $u=\sqrt{x}$ and calculate the value of $du$ by differentiating $u$ with respect to $x$. After finding the value of $du$, we will substitute the values we have in the integration of the given equation. After substituting the values and simplifying the obtained equation we can use the formula $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C$ to get the result which is in terms of $u$. To get the answer in terms of $x$ we will again substitute the value of $u$ which is $u=\sqrt{x}$ in the integration value.

Complete step-by-step answer:
Given that, $\sqrt{\dfrac{2-x}{x}}$
Integrating the given equation with respect to $x$, then we have $\int{\sqrt{\dfrac{2-x}{x}}}dx$
Let us taking the substitution $u=\sqrt{x}$$\Rightarrow {{u}^{2}}=x$
Differentiating the value of $u$ with respect to $x$, then we will get
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( \sqrt{x} \right)$
We can write square root of any function as the $\dfrac{1}{2}$ power of that function, then
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( {{x}^{\dfrac{1}{2}}} \right)$
We have the differentiation value of ${{x}^{n}}$ as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
$\begin{align}
  & \therefore \dfrac{du}{dx}=\dfrac{d}{dx}\left( {{x}^{\dfrac{1}{2}}} \right)=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}} \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{1}{2}{{x}^{\dfrac{1-2}{2}}} \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}} \\
\end{align}$
We know that ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$, then we will get
$\begin{align}
  & \dfrac{du}{dx}=\dfrac{1}{2}\dfrac{1}{{{x}^{\dfrac{1}{2}}}} \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{1}{2\sqrt{x}} \\
\end{align}$
From the value of $\dfrac{du}{dx}$ we can write $\dfrac{dx}{\sqrt{x}}=2du$
We have values $\dfrac{dx}{\sqrt{x}}=2du$, $u=\sqrt{x}$ and ${{u}^{2}}=x$. Substituting all those values in the integration, then we will get
$\begin{align}
  & \int{\sqrt{\dfrac{2-x}{x}}}dx=\int{\dfrac{\sqrt{2-x}}{\sqrt{x}}dx} \\
 & \Rightarrow \int{\sqrt{\dfrac{2-x}{x}}}dx=\int{2\sqrt{2-{{u}^{2}}}}du \\
 & \Rightarrow \int{\sqrt{\dfrac{2-x}{x}}}dx=2\int{\sqrt{2-{{u}^{2}}}}du \\
\end{align}$
Now we have the formula $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C$, then
$\int{\sqrt{{{\left( \sqrt{2} \right)}^{2}}-{{u}^{2}}}}du=\dfrac{u}{2}\sqrt{2-{{u}^{2}}}+\dfrac{{{\left( \sqrt{2} \right)}^{2}}}{2}{{\sin }^{-1}}\dfrac{u}{\sqrt{2}}+C$
Substituting the value of $u=\sqrt{x}$ in the above equation, then we will get
$\begin{align}
  & \int{\sqrt{\dfrac{2-x}{x}}}dx=\dfrac{\sqrt{x}}{2}\sqrt{2-{{\left( \sqrt{x} \right)}^{2}}}+\dfrac{2}{2}{{\sin }^{-1}}\dfrac{\sqrt{x}}{\sqrt{2}}+C \\
 & \Rightarrow \int{\sqrt{\dfrac{2-x}{x}}}dx=\dfrac{\sqrt{x}}{2}\sqrt{2-x}+{{\sin }^{-1}}\left( \sqrt{\dfrac{x}{2}} \right)+C \\
\end{align}$

Note: We can solve this problem not only by substituting $u=\sqrt{x}$ but also, we can solve this problem by taking the substitution $u=\dfrac{\sqrt{x}}{\sqrt{2-x}}$. If the take the substitution $u=\dfrac{\sqrt{x}}{\sqrt{2-x}}$ you need to work hard to get the value of $\dfrac{du}{dx}$ and we will get the integration value in terms of ${{\tan }^{-1}}$ function.