Question

How do you integrate the function $\arctan \left( \sqrt{x} \right)$?

Answer 1

We start solving the problem by recalling the fact that $\arctan \left( x \right)={{\tan }^{-1}}\left( x \right)$. We then assume $x={{y}^{2}}$ and then apply a differential on both sides to get the relation between $dy$ and $dx$. We then substitute all these results in the integration and then recall the integration by parts of function $\int{f\left( x \right)g\left( x \right)dx}$ as $\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( \dfrac{d\left( f\left( x \right) \right)}{dx}\int{g\left( x \right)dx} \right)dx}$. We then make use of the fact that $\int{xdx}=\dfrac{{{x}^{2}}}{2}+C$, $\dfrac{d\left( {{\tan }^{-1}}\left( x \right) \right)}{dx}=\dfrac{1}{1+{{x}^{2}}}$ and $\int{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}\left( x \right)+C$ to proceed through the problem. We then make the necessary calculations to get the required answer of integration.

Complete step-by-step solution:
According to the problem, we are asked to integrate the given function $\arctan \left( \sqrt{x} \right)$.
We know that $\arctan \left( x \right)={{\tan }^{-1}}\left( x \right)$. So, we get $\arctan \left( \sqrt{x} \right)={{\tan }^{-1}}\left( \sqrt{x} \right)$.
Now, let us assume $I=\int{{{\tan }^{-1}}\left( \sqrt{x} \right)dx}$ ---(1).
Let us assume $x={{y}^{2}}$ ---(2).
Now, let us apply a differential on both sides of equation (2).
$\Rightarrow d\left( x \right)=d\left( {{y}^{2}} \right)$ ---(3).
We know that $d\left( {{x}^{2}} \right)=2xdx$. Let us use this result in equation (3).
$\Rightarrow dx=2ydy$ ---(4).
Let us substitute equations (2) and (4) in equation (1).
$\Rightarrow I=\int{{{\tan }^{-1}}\left( \sqrt{{{y}^{2}}} \right)\left( 2y \right)dy}$.
$\Rightarrow I=2\int{y{{\tan }^{-1}}\left( y \right)dy}$ ---(5).
Whenever we get the integration which resembles $\int{f\left( x \right)g\left( x \right)dx}$, then we make use of by parts integration as $\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( \dfrac{d\left( f\left( x \right) \right)}{dx}\int{g\left( x \right)dx} \right)dx}$. Let us use this result in equation (5).
$\Rightarrow I=2\left( {{\tan }^{-1}}\left( y \right)\int{ydy}-\int{\left( \dfrac{d\left( {{\tan }^{-1}}\left( y \right) \right)}{dy}\int{ydy} \right)dy} \right)$ ---(6).
We know that $\int{xdx}=\dfrac{{{x}^{2}}}{2}+C$ and $\dfrac{d\left( {{\tan }^{-1}}\left( x \right) \right)}{dx}=\dfrac{1}{1+{{x}^{2}}}$. Let us use this result in equations (6).
$\Rightarrow I=2\left( {{\tan }^{-1}}\left( y \right)\left( \dfrac{{{y}^{2}}}{2} \right)-\int{\left( \left( \dfrac{1}{1+{{y}^{2}}} \right)\times \left( \dfrac{{{y}^{2}}}{2} \right) \right)dy} \right)$.
$\Rightarrow I=2\left( \left( \dfrac{{{y}^{2}}{{\tan }^{-1}}\left( y \right)}{2} \right)-\dfrac{1}{2}\int{\dfrac{{{y}^{2}}}{1+{{y}^{2}}}dy} \right)$.
$\Rightarrow I=2\left( \left( \dfrac{{{y}^{2}}{{\tan }^{-1}}\left( y \right)}{2} \right)-\dfrac{1}{2}\int{\dfrac{{{y}^{2}}+1-1}{1+{{y}^{2}}}dy} \right)$.
$\Rightarrow I=2\left( \left( \dfrac{{{y}^{2}}{{\tan }^{-1}}\left( y \right)}{2} \right)-\dfrac{1}{2}\int{1-\dfrac{1}{1+{{y}^{2}}}dy} \right)$ ---(7).
We know that $\int{adx}=ax+C$ and $\int{\dfrac{1}{1+{{x}^{2}}}dx}={{\tan }^{-1}}\left( x \right)+C$. Let us use these results in equation (7).
$\Rightarrow I=2\left( \left( \dfrac{{{y}^{2}}{{\tan }^{-1}}\left( y \right)}{2} \right)-\dfrac{1}{2}\left( y-{{\tan }^{-1}}\left( y \right) \right) \right)+C$.
$\Rightarrow I={{y}^{2}}{{\tan }^{-1}}\left( y \right)-y+{{\tan }^{-1}}\left( y \right)+C$ ---(8).
From equation (2), we have ${{y}^{2}}=x\Leftrightarrow y=\sqrt{x}$. Let us substitute this in equation (8).
$\Rightarrow I=x{{\tan }^{-1}}\left( \sqrt{x} \right)+{{\tan }^{-1}}\left( \sqrt{x} \right)-\sqrt{x}+C$.
$\therefore$ We have found the integration of $\arctan \left( \sqrt{x} \right)$ as $x{{\tan }^{-1}}\left( \sqrt{x} \right)+{{\tan }^{-1}}\left( \sqrt{x} \right)-\sqrt{x}+C$.

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. We should not forget to add constants of integration while solving the problems related to indefinite integrals. We should not forget to replace the Y with the equivalent function of x after equation (8) as this is the most common mistake done by students. Similarly, we can expect problems to find the value of $I=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \sqrt{x} \right)dx}$.