Question

Evaluate the following expression $\int{\sqrt{\tan x}dx}$.

Answer 1

Hint: In order to solve this question, we will first consider $\tan x={{t}^{2}}$ and then we will simplify it to form an easily integrable form and then we will integrate it to get our answer. While solving this question, we need to remember that $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}}$ and $\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{1}{2a}\ln }\left| \dfrac{x-a}{x+a} \right|$, By using these, we can solve this question.
Complete step-by-step answer:
In this question, we have asked to find the integral of \[\sqrt{\tan x}\]. To solve this question, let us consider, $I=\int{\sqrt{\tan x}}dx$ and $\sqrt{\tan x}=t$. So, we can say $d\left( \sqrt{\tan x} \right)=d\left( t \right)$ and we get, $\dfrac{{{\sec }^{2}}x}{2\sqrt{\tan x}}dx=dt$. Now, we know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x=1+{{t}^{4}}$. Therefore, we get, $\dfrac{1+{{t}^{4}}}{2t}dx=dt$, which can be written as, $dx=\dfrac{2t}{1+{{t}^{4}}}dt$. Therefore, we get the integral as,
$\begin{align}
  & I=\int{\dfrac{2{{t}^{2}}}{1+{{t}^{4}}}}dt \\
 & I=\int{\dfrac{t.2t}{1+{{t}^{4}}}}dt \\
\end{align}$
And we can further write it as,
$I=\int{\dfrac{{{t}^{2}}+{{t}^{2}}}{1+{{t}^{4}}}}dt$
Now, we will add and subtract 1 from the numerator. So, we get,
$\begin{align}
  & I=\int{\dfrac{{{t}^{2}}+1+{{t}^{2}}-1}{1+{{t}^{4}}}}dt \\
 & I=\int{\left( \dfrac{{{t}^{2}}+1}{1+{{t}^{4}}}+\dfrac{{{t}^{2}}-1}{1+{{t}^{4}}} \right)}dt \\
\end{align}$
Now, we will take ${{t}^{2}}$ common from both the numerator and the denominator. So, we get,
$I=\int{\left( \dfrac{1+\dfrac{1}{{{t}^{2}}}}{{{t}^{2}}+\dfrac{1}{{{t}^{2}}}}+\dfrac{1-\dfrac{1}{{{t}^{2}}}}{{{t}^{2}}+\dfrac{1}{{{t}^{2}}}} \right)}dt$
Now, we will add and subtract 2 from the denominator. So, we get,
$I=\int{\left( \dfrac{1+\dfrac{1}{{{t}^{2}}}}{{{t}^{2}}+\dfrac{1}{{{t}^{2}}}+2-2}+\dfrac{1-\dfrac{1}{{{t}^{2}}}}{{{t}^{2}}+\dfrac{1}{{{t}^{2}}}+2-2} \right)}dt$
Now, we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, for a = t and $b=\dfrac{1}{t}$, we get, ${{\left( t-\dfrac{1}{t} \right)}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2$ and ${{\left( t+\dfrac{1}{t} \right)}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}+2$. Therefore, we can write the first and the second term of the integral as,
$I=\int{\left( \dfrac{1+\dfrac{1}{{{t}^{2}}}}{{{\left( t-\dfrac{1}{t} \right)}^{2}}+2}+\dfrac{1-\dfrac{1}{{{t}^{2}}}}{{{\left( t+\dfrac{1}{t} \right)}^{2}}-2} \right)}dt$
And, it can be further written as,
$I=\int{\dfrac{1+\dfrac{1}{{{t}^{2}}}}{{{\left( t-\dfrac{1}{t} \right)}^{2}}+2}dt+\dfrac{1-\dfrac{1}{{{t}^{2}}}}{{{\left( t+\dfrac{1}{t} \right)}^{2}}-2}dt}$
Now, we will consider, \[\int{\dfrac{1+\dfrac{1}{{{t}^{2}}}}{{{\left( t-\dfrac{1}{t} \right)}^{2}}+2}dt}\] as A and \[\int{\dfrac{1-\dfrac{1}{{{t}^{2}}}}{{{\left( t+\dfrac{1}{t} \right)}^{2}}-2}}dt\] as B. So, we can write I as,
I = A + B……… (i)
Now, we will simplify A and B individually to get the value of I. So, we can write,
\[A=\int{\dfrac{1+\dfrac{1}{{{t}^{2}}}}{{{\left( t-\dfrac{1}{t} \right)}^{2}}+2}dt}\] and \[B=\int{\dfrac{1-\dfrac{1}{{{t}^{2}}}}{{{\left( t+\dfrac{1}{t} \right)}^{2}}-2}}dt\]
Now, consider \[\left( t-\dfrac{1}{t} \right)=u\] and \[\left( t+\dfrac{1}{t} \right)=v\]. So, we can write, $\left( 1+\dfrac{1}{{{t}^{2}}} \right)dt=du$ and $\left( 1-\dfrac{1}{{{t}^{2}}} \right)dt=dv$
Therefore, we can write A and B as,
 \[A=\int{\dfrac{1}{{{u}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}du}\] and \[B=\int{\dfrac{1}{{{v}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}dv}\]
Now, we know that $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}}$. So, for x = u and $a=\sqrt{2}$, we can write,
\[\int{\dfrac{1}{{{u}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}du}=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{u}{\sqrt{2}} \right)\]. We also know that $\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{1}{2a}\ln }\left| \dfrac{x-a}{x+a} \right|$. So, for, x = v and $a=\sqrt{2}$, we can write, $\int{\dfrac{1}{{{v}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}dv=\dfrac{1}{2\sqrt{2}}\ln }\left| \dfrac{v-\sqrt{2}}{v+\sqrt{2}} \right|$
Therefore, we can write A and B as,
$A=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{u}{\sqrt{2}} \right)+c'$ and $B=\dfrac{1}{2\sqrt{2}}\ln \left| \dfrac{v-\sqrt{2}}{v+\sqrt{2}} \right|+c''$
Now, we will put the values of u and v, that is, \[u=\left( t-\dfrac{1}{t} \right)\] and \[v=\left( t+\dfrac{1}{t} \right)\] in A and B. So, we get,
 $A=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{t-\dfrac{1}{t}}{\sqrt{2}} \right)+c'$ and $B=\dfrac{1}{2\sqrt{2}}\ln \left| \dfrac{t+\dfrac{1}{t}-\sqrt{2}}{t+\dfrac{1}{t}+\sqrt{2}} \right|+c''$
Now, we will put the values of A and B in equation (i), so we get,
$I=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{t-\dfrac{1}{t}}{\sqrt{2}} \right)+\dfrac{1}{2\sqrt{2}}\ln \left| \dfrac{t+\dfrac{1}{t}-\sqrt{2}}{t+\dfrac{1}{t}+\sqrt{2}} \right|+c'+c''$
Now, we will put the Values of t, that is, \[\sqrt{\tan x}\] and c = c’+ c’’. So, we get I as,
$I=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{\sqrt{\tan x}-\dfrac{1}{\sqrt{\tan x}}}{\sqrt{2}} \right)+\dfrac{1}{2\sqrt{2}}\ln \left| \dfrac{\sqrt{\tan x}+\dfrac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\dfrac{1}{\sqrt{\tan x}}+\sqrt{2}} \right|+c$
Hence, we can say that integral of \[\sqrt{\tan x}\] is $\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{\sqrt{\tan x}-\dfrac{1}{\sqrt{\tan x}}}{\sqrt{2}} \right)+\dfrac{1}{2\sqrt{2}}\ln \left| \dfrac{\sqrt{\tan x}+\dfrac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\dfrac{1}{\sqrt{\tan x}}+\sqrt{2}} \right|+c$

Note: While solving this question, there are high chances of calculation mistakes as this question contains a lot of calculation. Also, we need to remember a few derivatives like, $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ and $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$ and $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}$. Also, we have to remember the integrals of $\dfrac{1}{{{x}^{2}}-{{a}^{2}}}$ and $\dfrac{1}{{{x}^{2}}+{{a}^{2}}}$ as well. The possibility of mistake here is interchanging the above two integration formulas and getting the wrong answer.