Question

How do you integrate $\int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}}$ using trigonometric substitutions?

Answer 1

Hint: In the problem they have asked to use the trigonometric substitutions, by looking the given equation we will use the trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$. So, we will take the substitution ${{x}^{2}}=9{{\sec }^{2}}y$. From this equation we will calculate the value of $x$ and $dx$ in terms of $y$ and $dy$, by using the algebraic formulas and differentiation formulas. Now we will convert the given equation in terms of $y$ and $dy$ by substituting the value of $x$ and $dx$ in the given equation. Now we will simplify the obtained equation. After integrating the above value, we will get the result in terms of $y$, but we need to calculate the result in terms of $x$, so we will calculate the value of $y$ from our assumption i.e., ${{x}^{2}}=9{{\sec }^{2}}y$ and substitute it in the obtained equation to get the result.

Complete step-by-step solution:
Given that, $\int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}}$.
Let us take the substitution ${{x}^{2}}=9{{\sec }^{2}}y$.
From the above substitution the value of $x$ can be obtained by taking square root on both sides of the equation, then
$\begin{align}
  & \sqrt{{{x}^{2}}}=\sqrt{9{{\sec }^{2}}y} \\
 & \Rightarrow x=3\sec y....\left( \text{i} \right) \\
\end{align}$
Differentiating the above value, then we will get
$dx=3\sec y\tan y.dy$
Now substituting the all values, we have in the given equation, then we will get
$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}dx}=\int{\dfrac{1}{\sqrt{9{{\sec }^{2}}y-9}}3\sec y\tan ydy}$
Taking $9$ common in the square root function, then we will get
$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}dx}=\int{\dfrac{3\sec y\tan y}{\sqrt{9\left( {{\sec }^{2}}y-1 \right)}}dy}$
From the trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, we have the value ${{\sec }^{2}}y-1={{\tan }^{2}}y$, then the above equation modified as
$\begin{align}
  & \Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}dx}=\int{\dfrac{3\sec y\tan y}{\sqrt{9{{\tan }^{2}}y}}dy} \\
 & \Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}dx}=\int{\dfrac{3\sec y\tan y}{3\tan y}dy} \\
\end{align}$
Cancelling the terms which are in both numerator and denominator, then we will get
$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}dx}=\int{\sec ydy}$
We have the integration formulas $\int{\sec ydy}=\log \left| \sec y+\tan y \right|+C$ then we will get
$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}dx}=\log \left| \sec y+\tan y \right|+C$
From equation $\left( \text{i} \right)$, we have
$\begin{align}
  & x=3\sec y \\
 & \Rightarrow \sec y=\dfrac{x}{3}...\left( \text{ii} \right) \\
\end{align}$
From the trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, then value of $\tan y$ will be
$\begin{align}
  & {{\tan }^{2}}y={{\sec }^{2}}y-1 \\
 & \Rightarrow \tan y=\sqrt{{{\left( \dfrac{x}{3} \right)}^{2}}-1}.....\left( \text{iii} \right) \\
\end{align}$
From equations $\left( \text{ii} \right)$, $\left( \text{iii} \right)$ the value of $\int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}}$ is
$\Rightarrow \int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}}dx=\log \left| \dfrac{x}{3}+\sqrt{{{\left( \dfrac{x}{3} \right)}^{2}}-1} \right|+C$
Hence the value of $\int{\dfrac{1}{\sqrt{{{x}^{2}}-9}}}dx$ is $\log \left| \dfrac{x}{3}+\dfrac{1}{3}\sqrt{{{x}^{2}}-9} \right|+C$.

Note: In the problem they have specially mentioned to use the trigonometric substitution, so we have followed the above procedure otherwise we have direct formula for this problem i.e., $\int{\dfrac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$. We will convert the given equation in the above form and use this formula to get the result.