Question

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

Answer 1

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.

& \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{aligned}$$ We know that the value of $${{\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{aligned} & \Rightarrow I=\dfrac{\pi }{4}-0 \\\ & \Rightarrow I=\dfrac{\pi }{4}.....(5) \\\ \end{aligned}$$ From equation (5), it is clear that the value of I is equal to $$\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.