Question

Integrate the following expression with respect to x:
$\int{\sqrt{\tan x}dx}$

Answer 1

Hint: First we will substitute tanx and then we will convert the trigonometric form into algebraic form and after we will separate the expression into two and then we try to solve them separately by using some algebraic formulas and then we will again substitute and try to convert them into a form whose integration is known to us.

Complete step-by-step answer:

Let’s start by substituting tanx with ${{t}^{2}}$ we get,

$\begin{align}

  & \tan x={{t}^{2}} \\

 & \Rightarrow {{\sec }^{2}}xdx=2tdt \\

 & \Rightarrow dx=\dfrac{2tdt}{{{\sec }^{2}}x} \\

 & \Rightarrow dx=\dfrac{2tdt}{1+{{\tan }^{2}}x} \\

 & \Rightarrow dx=\dfrac{2tdt}{1+{{t}^{4}}} \\

\end{align}$

Now we will substitute the value of tanx and dx in $\int{\sqrt{\tan x}dx}$ we get,

$\begin{align}

  & \int{t\times \dfrac{2tdt}{1+{{t}^{4}}}} \\

 & \int{\dfrac{2{{t}^{2}}dt}{1+{{t}^{4}}}} \\

 & \int{\dfrac{\left( {{t}^{2}}+1+{{t}^{2}}-1 \right)dt}{1+{{t}^{4}}}} \\

 & \int{\dfrac{\left( {{t}^{2}}+1 \right)dt}{1+{{t}^{4}}}}+\int{\dfrac{\left( {{t}^{2}}-1 \right)dt}{1+{{t}^{4}}}} \\

\end{align}$

Now we are going to divide both the numerator and denominator by ${{t}^{2}}$ in both the expression,

$\int{\dfrac{\left( \dfrac{1}{{{t}^{2}}}+1 \right)dt}{\dfrac{1}{{{t}^{2}}}+{{t}^{2}}}}+\int{\dfrac{\left( -\dfrac{1}{{{t}^{2}}}+1 \right)dt}{\dfrac{1}{{{t}^{2}}}+{{t}^{2}}}}$

Now we are going to use the formula:

$\begin{align}

  & {{\left( a-\dfrac{1}{a} \right)}^{2}}={{a}^{2}}+\dfrac{1}{{{a}^{2}}}-2 \\

 & {{\left( a+\dfrac{1}{a} \right)}^{2}}={{a}^{2}}+\dfrac{1}{{{a}^{2}}}+2 \\

\end{align}$

Using this formula we get,

$\int{\dfrac{\left( \dfrac{1}{{{t}^{2}}}+1 \right)dt}{{{\left( t-\dfrac{1}{t} \right)}^{2}}+2}}+\int{\dfrac{\left( -\dfrac{1}{{{t}^{2}}}+1 \right)dt}{{{\left( t+\dfrac{1}{t} \right)}^{2}}-2}}$

Now substituting $t-\dfrac{1}{t}=s\text{ and }t+\dfrac{1}{t}=r$ we get,

$\int{\dfrac{ds}{{{s}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}+}\int{\dfrac{dr}{{{r}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}}............(1)$

Now we know that,

$\begin{align}

  & \int{\dfrac{dx}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}+c} \\

 & \int{\dfrac{dx}{{{x}^{2}}-{{a}^{2}}}}=\dfrac{1}{2a}\log \left| \dfrac{x-a}{x+a} \right|+c \\

\end{align}$

Now using the above two formula in equation (1) we get,

$\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{s}{\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\log \left| \dfrac{r-a}{r+a} \right|+c$

Now we have integrated the given expression, so we can substitute back the terms.
$\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{t-\dfrac{1}{t}}{\sqrt{2}}+\dfrac{1}{2\sqrt{2}}\log \left| \dfrac{t+\dfrac{1}{t}-a}{t+\dfrac{1}{t}+a} \right|+c$

Here t = $\sqrt{\tan x}$ .

Hence this is the final answer to this question.

Note: There many trigonometric formulas that we have used and the method of substitution is very important as many students stuck at that point and they don’t know what to substitute, so observe the answer carefully and the formulas of integration that we have used should be remembered so that there is no problem in understanding. This question tests most of the concept of integration and how to use algebraic formulas to proceed further.