Question

Evaluate the following integral \[\int\limits_{0}^{3}{\dfrac{dx}{9+{{x}^{2}}}}\].

Answer 1

Hint: We know that the value of \[\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}\] is equal to \[\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\]. Let us assume the value of \[\int\limits_{0}^{3}{\dfrac{dx}{9+{{x}^{2}}}}\] is equal to I. Let us assume this as equation (1). We know that 9 can be written as \[{{3}^{2}}\]. Now we will substitute \[{{3}^{2}}\] in equation (1). Now by using the formula, \[\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\] we will find the value of \[\int\limits_{0}^{3}{\dfrac{dx}{9+{{x}^{2}}}}\].

Complete step by step answer:
Before solving we should know that \[\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\].
From the question, it was given that to evaluate \[\int\limits_{0}^{3}{\dfrac{dx}{9+{{x}^{2}}}}\].
Let us assume the value of \[\int\limits_{0}^{3}{\dfrac{dx}{9+{{x}^{2}}}}\] is equal to I.
\[\Rightarrow I=\int\limits_{0}^{3}{\dfrac{dx}{9+{{x}^{2}}}}....(1)\]
We know that 9 can be written as \[{{3}^{2}}\].
 \[\Rightarrow I=\int\limits_{0}^{3}{\dfrac{dx}{{{3}^{2}}+{{x}^{2}}}}\]
We know that \[\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\].
Now we will apply this formula to find the value of I.
\[\begin{align}
  & \Rightarrow I=\left[ {{\tan }^{-1}}\left( \dfrac{x}{3} \right) \right]_{0}^{3} \\
 & \Rightarrow I={{\tan }^{-1}}\left( \dfrac{3}{3} \right)-{{\tan }^{-1}}\left( \dfrac{0}{3} \right) \\
 & \Rightarrow I={{\tan }^{-1}}1-{{\tan }^{-1}}0.....(2) \\
\end{align}\]
We know that the value of \[{{\tan }^{-1}}1\] is equal to \[\dfrac{\pi }{4}\].
Let us consider
\[\Rightarrow {{\tan }^{-1}}1=\dfrac{\pi }{4}.....(3)\]
We also know that the value of \[{{\tan }^{-1}}0\] is equal to 0.
Let us consider
\[\Rightarrow {{\tan }^{-1}}0=0.....(4)\]
Now we will substitute equation (3) and equation (4) in equation (2). Then we get
\[\begin{align}
  & \Rightarrow I=\dfrac{\pi }{4}-0 \\
 & \Rightarrow I=\dfrac{\pi }{4}.....(5) \\
\end{align}\]
From equation (5), it is clear that the value of I is equal to \[\dfrac{\pi }{4}\].
So, we can say that the value of \[\int\limits_{0}^{3}{\dfrac{dx}{9+{{x}^{2}}}}\] is equal to \[\dfrac{\pi }{4}\].


Note:
Students may have a misconception that the value of \[\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}\] is equal to \[\dfrac{1}{2a}\log \left( \dfrac{a+x}{a-x} \right)\]. But we know that the value of \[\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}\] is equal to \[\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)\]. So, if this misconception is followed, then we will definitely get the wrong answer. So, students should avoid this misconception. Students should also avoid calculation mistakes during solving this problem. If a small mistake is done, the final answer may get interrupted. So, calculation should be done in a careful manner.