Question

Evaluate the following integral: \[\int{{{e}^{{{\tan }^{-1}}x}}\left( 1+x+{{x}^{2}} \right)d\left( {{\cot }^{-1}}x \right)}\]

Answer 1

Hint: We will first start by simplifying the given integral and we will do that by substituting $t={{\tan }^{-1}}x$ , then we will apply trigonometric property that is ${{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}$ , this will help us in simplifying the term $d\left( {{\cot }^{-1}}x \right)$ , we will proceed further by solving the integral using integration by parts and then again we will re-substitute the given values and get the answer.

Complete step by step answer:
Let $t={{\tan }^{-1}}x$ , taking inverse of tan to the other side we will get the following:
$x=\tan t$ .
Now again we know that \[{{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}\]
On differentiating both the sides of the equation we will get: \[d({{\cot }^{-1}}x)+d({{\tan }^{-1}}x)=0\Rightarrow d({{\cot }^{-1}}x)=-d({{\tan }^{-1}}x).............\text{Equation 1}\text{.}\]
Now at the start of the question we have assumed $t={{\tan }^{-1}}x$ , now on differentiating it, we will get: $dt=d\left( {{\tan }^{-1}}x \right)$ . Now we will substitute this in equation 1 and therefore, we will get: $d\left( {{\cot }^{-1}}x \right)=-\left( dt \right)$
Now let’s take the given integral and let \[I=\int{{{e}^{{{\tan }^{-1}}x}}\left( 1+x+{{x}^{2}} \right)d\left( {{\cot }^{-1}}x \right)}\]
Now let’s put the value of $d\left( {{\cot }^{-1}}x \right)=-\left( dt \right)$
Therefore \[I=\int{{{e}^{{{\tan }^{-1}}x}}\left( 1+x+{{x}^{2}} \right)d\left( {{\cot }^{-1}}x \right)}\Rightarrow I=-\int{{{e}^{{{\tan }^{-1}}x}}\left( 1+x+{{x}^{2}} \right)dt}\text{ }..........\text{Equation 2}\text{.}\]
We have assumed in the starting of our solution that $t={{\tan }^{-1}}x\Rightarrow x=\tan t$ , now putting this value of x in equation 2.
We will have: \[I=-\int{{{e}^{{{\tan }^{-1}}x}}\left( 1+x+{{x}^{2}} \right)dt}\text{ }\Rightarrow I=-\int{{{e}^{t}}\left( 1+\tan t+{{\tan }^{2}}t \right)}dt\] , we will now expand this integral as we know that: $\int{\left( f\left( x \right)+g\left( x \right) \right)}dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx}}$
Therefore,
\[I=-\int{{{e}^{t}}\left( 1+\tan t+{{\tan }^{2}}t \right)}dt\Rightarrow I=-\left[ \int{{{e}^{t}}dt+}\int{{{e}^{t}}\tan tdt+\int{{{e}^{t}}{{\tan }^{2}}tdt}} \right].................\text{ Equation 3}\text{.}\]
Let: $I=-\left[ {{I}_{1}}+{{I}_{2}}+{{I}_{3}} \right]$ , where :
\[\begin{align}
  & {{I}_{1}}=\int{{{e}^{t}}dt} \\
 & {{I}_{2}}=\int{{{e}^{t}}\tan tdt} \\
 & {{I}_{3}}=\int{{{e}^{t}}{{\tan }^{2}}tdt} \\
\end{align}\]
We will put ${{I}_{1}}$ and ${{I}_{3}}$ as it is and solve ${{I}_{2}}$ by the method of Integration by parts, This formula is used for integrating the product of two functions, in this method we find the integrals by reducing them into standard forms. The integral of the two functions is taken, by considering the left term as first function and second term as the second function. There is a method called ILATE rule which helps in deciding the first and second function. So, the order for choosing the first function is Inverse, Logarithm, Algebra, Trigonometry and Exponent. The formula for integrating by parts is given as follows:
$\int{u\dfrac{dv}{dx}dx=uv-\int{v\dfrac{du}{dx}dx}}$

We have, \[{{I}_{2}}=\int{{{e}^{t}}\tan tdt}\] ,
Let us take $u=\tan t\text{ and v=}{{\text{e}}^{t}}$ as per the ILATE rule and then applying integration by parts methods as mentioned above, therefore
\[{{I}_{2}}=\int{\tan t{{e}^{t}}dt}\]
${{I}_{2}}=\int{\tan t.{e^{t}}dt=\tan t{{e}^{t}}-\int{{{e}^{t}}\dfrac{d\left( \tan t \right)}{dt}dt}}$
\[{{I}_{2}}=\int{{{e}^{t}}\tan tdt}=\left\{ \tan t{{e}^{t}}-\int{{{\sec }^{2}}t{{e}^{t}}dt} \right\}\]
We know that ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ , Therefore: \[\Rightarrow {{I}_{2}}=\tan t{{e}^{t}}-\int{\left( 1+{{\tan }^{2}}t \right){{e}^{t}}dt}\text{ }........\text{Equation 4}\text{.}\]
Putting this value of ${{I}_{2}}$ in Equation 3 and letting the values of ${{I}_{1}}$ and ${{I}_{3}}$ as it is that is : \[{{I}_{1}}=\int{{{e}^{t}}dt}\] and \[{{I}_{3}}=\int{{{e}^{t}}{{\tan }^{2}}tdt}\]. We will first write $I=-\left[ {{I}_{1}}+{{I}_{2}}+{{I}_{3}} \right]$ as $I=-\left[ {{I}_{1}}+{{I}_{3}}+{{I}_{2}} \right]$
Therefore equation 3 becomes: \[I=-\left[ \int{{{e}^{t}}dt+}\int{{{e}^{t}}\tan tdt+\int{{{e}^{t}}{{\tan }^{2}}tdt}} \right]\Rightarrow I=-\left[ \int{{{e}^{t}}\left( 1+{{\tan }^{2}}t \right)dt}+\int{{{e}^{t}}\tan tdt} \right]\]
Putting the value of ${{I}_{2}}$ from equation 4 in the above equation we get:
\[\begin{align}
  & I=-\left[ \int{{{e}^{t}}\left( 1+{{\tan }^{2}}t \right)dt}+\tan t{{e}^{t}}-\int{\left( 1+{{\tan }^{2}}t \right){{e}^{t}}dt} \right] \\
 & \Rightarrow I=-\tan t{{e}^{t}}\Rightarrow I=-{{e}^{t}}\tan t \\
\end{align}\]
Now after re-substituting $t={{\tan }^{-1}}x$ , we will get:
\[I=-{{e}^{t}}\tan t\Rightarrow I=-{{e}^{{{\tan }^{-1}}x}}\tan ({{\tan }^{-1}}x)\]
We already know that according to the trigonometric property: $\tan \left( {{\tan }^{-1}}\theta \right)=\theta $
Therefore the value of $I$ becomes: $I=-{{e}^{{{\tan }^{-1}}x}}x\Rightarrow I=-x{{e}^{{{\tan }^{-1}}x}}$

Hence, the answer is: $-x{{e}^{{{\tan }^{-1}}x}}$.

Note: In these types of questions a lot of trigonometric properties comes in hand, so you must know them or know how to derive them. You could also do this integration by solving each and every part of $I$ , but that would become way too much long and chances of mistakes are higher, and hence try to convert the terms that can cancel each other, this will make our calculation easy and simple.