Question

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

Answer 1

In the given question, we have been asked to integrate the following function. In order to solve the question, we integrate the numerical by following the trigonometric substitution method. After solving the integration and depending on the resultant integration we need to integrate further, we will substitute one of the trigonometric expressions to simplify the given integration further.

Complete step by step solution:
We have given,
$\Rightarrow \int{\dfrac{1}{x-\sqrt{9+{{x}^{2}}}}dx}$
Multiply the numerator and denominator of the given expression by the conjugate of the denominator, we get
$\Rightarrow \int{\dfrac{1}{x-\sqrt{9+{{x}^{2}}}}\times \dfrac{x+\sqrt{9+{{x}^{2}}}}{x+\sqrt{9+{{x}^{2}}}}dx}$
As we know that, $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Simplifying the above expression, we get
$\Rightarrow \int{\dfrac{x+\sqrt{9+{{x}^{2}}}}{{{x}^{2}}-\left( 9+{{x}^{2}} \right)}dx}$
Simplifying this by taking out the constant part, we get
$\Rightarrow -\dfrac{1}{9}\int{\left( x+\sqrt{9+{{x}^{2}}} \right)dx}$
$\Rightarrow -\dfrac{1}{9}\int{xdx-\dfrac{1}{9}\int{\sqrt{9+{{x}^{2}}}dx}}$
Integrating $\int{xdx}$, we obtain
$\Rightarrow -\dfrac{1}{18}{{x}^{2}}-\dfrac{1}{9}\int{\sqrt{9+{{x}^{2}}}dx}$
Now,
Let J = $\int{\sqrt{9+{{x}^{2}}}dx}$.
Now substitute $x=3\tan \theta$
Thus,
$x=3\tan \theta$
$dx=3{{\sec }^{2}}\theta d\theta$
Substituting these values in the above expression, we get
$J=\int{\sqrt{9+{{x}^{2}}}dx}=\int{\sqrt{9+9{{\tan }^{2}}\theta }\left( {{\sec }^{2}}\theta d\theta \right)=}9\int{\sqrt{1+{{\tan }^{2}}\theta }\left( {{\sec }^{2}}\theta d\theta \right)}$
As we know that, $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$
Applying this in the above expression, we get
$\Rightarrow J=9\int{\sqrt{{{\sec }^{2}}\theta }\left( {{\sec }^{2}}\theta d\theta \right)}=9\int{\sec \theta }\left( {{\sec }^{2}}\theta d\theta \right)=9\int{{{\sec }^{3}}\theta }d\theta$
$\Rightarrow J=9\int{{{\sec }^{3}}\theta }d\theta$
Now, solving the resultant integration by parts; i.e.
$\int{udv=uv-\int{vdu}}$
We have,
$\Rightarrow 9\int{{{\sec }^{3}}\theta }d\theta$
Let k = $\int{{{\sec }^{3}}\theta }d\theta$
Here,
$\Rightarrow u=\sec \theta \ then\ \Rightarrow du=\sec \theta \tan \theta d\theta$
$\Rightarrow dv={{\sec }^{2}}\theta d\theta \ then\ \Rightarrow v=\tan \theta$
Thus, putting these values in the by-parts formula of integration, we get
$\Rightarrow k=\int{{{\sec }^{3}}\theta }d\theta =\sec \theta \tan \theta -\int{\sec \theta {{\tan }^{2}}\theta d\theta }$
Using the trigonometric rule i.e. ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$,
$\Rightarrow \sec \theta \tan \theta -\int{\sec \theta \left( {{\sec }^{2}}\theta -1 \right)d\theta }=\sec \theta \tan \theta -\int{{{\sec }^{3}}\theta d\theta +}\int{\sec \theta d\theta }$
Undo the substitution i.e. k = $\int{{{\sec }^{3}}\theta }d\theta$ in the above integral, we obtain
$\Rightarrow k=\sec \theta \tan \theta -k+\int{\sec \theta d\theta }$
Adding ‘k’ to both the sides of the equation, we get
$\Rightarrow 2k=\sec \theta \tan \theta +\int{\sec \theta d\theta }$
Now integration of $\int{\sec \theta d\theta }=\ln \left| \sec \theta +\tan \theta \right|$
Substitute in the above integral, we get
$\Rightarrow 2k=\sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right|$
Now, solving for the value of ‘k’, we get
Divide both the sides of the expression by 2, we get
$\Rightarrow k=\dfrac{\left( \sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \right)}{2}$
And we know that, $k=\int{{{\sec }^{3}}\theta }d\theta$
Therefore,
$\Rightarrow k=\int{{{\sec }^{3}}\theta d\theta =}\dfrac{\left( \sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \right)}{2}$
Returning to $J=9\int{{{\sec }^{3}}\theta }d\theta$,
Putting the value of $\int{{{\sec }^{3}}\theta d\theta =}\dfrac{\left( \sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \right)}{2}$ in $J=9\int{{{\sec }^{3}}\theta }d\theta$, we get
$\Rightarrow J=9\int{{{\sec }^{3}}\theta }d\theta =9\times \dfrac{\left( \sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \right)}{2}$
Simplifying the above expression, we get
$\Rightarrow J=\dfrac{9}{2}\left( \sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \right)$
Writing the above expression in form of $\tan \theta$, we get
$\Rightarrow J=\dfrac{9}{2}\left( \sqrt{{{\tan }^{2}}\theta +1}\left( \tan \theta \right) \right)+\ln \left| \sqrt{{{\tan }^{2}}\theta +1}+\tan \theta \right|$
Putting the value of $\tan \theta =\dfrac{x}{3}$, we get
$\Rightarrow J=\dfrac{9}{2}\left( \sqrt{\dfrac{{{x}^{2}}}{9}+1}\left( \dfrac{x}{3} \right)+\ln \left| \sqrt{\dfrac{{{x}^{2}}}{9}+1}+\left( \dfrac{x}{3} \right) \right| \right)$
Simplifying the above expression, we get
$\Rightarrow J=\dfrac{9}{2}\left( \sqrt{\dfrac{1}{9}\left( {{x}^{2}}+9 \right)}\left( \dfrac{x}{3} \right)+\ln \left| \sqrt{\dfrac{1}{9}\left( {{x}^{2}}+9 \right)}+\left( \dfrac{x}{3} \right) \right| \right)$
Solving the above, we get
$\Rightarrow J=\dfrac{9}{2}\left( \dfrac{x}{9}\sqrt{{{x}^{2}}+9}+\ln \left| \dfrac{1}{3}\left( \sqrt{{{x}^{2}}+9}+x \right) \right| \right)$
Solving for the each part individually, we get
$\Rightarrow J=\dfrac{x}{2}\sqrt{{{x}^{2}}+9}+\dfrac{9}{2}\ln \left| \sqrt{{{x}^{2}}+9}+x \right|-\dfrac{9}{2}\ln \left( 3 \right)$
Here, $\dfrac{9}{2}\ln \left( 3 \right)$ is a constant part and in the integral we write ‘c’ for it,
Substituting the value of ‘J’ in the integration we have solved first, we obtain
$\Rightarrow I=-\dfrac{1}{18}{{x}^{2}}-\dfrac{1}{9}\int{\sqrt{9+{{x}^{2}}}dx}$
$\Rightarrow I=-\dfrac{1}{18}{{x}^{2}}-\dfrac{1}{9}\left( \dfrac{x}{2}\sqrt{{{x}^{2}}+9}+\dfrac{9}{2}\ln \left| \sqrt{{{x}^{2}}+9}+x \right|-\dfrac{9}{2}\ln \left( 3 \right) \right)$
Simplifying the above expression, we get
$\Rightarrow I=-\dfrac{1}{18}{{x}^{2}}-\dfrac{1}{18}x\left( \sqrt{{{x}^{2}}+9} \right)-\dfrac{1}{2}\ln \left| \sqrt{{{x}^{2}}+9}+x \right|+C$

Therefore,
$\Rightarrow I=-\dfrac{{{x}^{2}}+x\sqrt{{{x}^{2}}+9}+9\ln \left| \sqrt{{{x}^{2}}+9}+x \right|}{18}+c$
Hence, it is the required answer.

Note: Students need to remember that if there is root in the denominator then we need to multiply the numerator and denominator by the conjugate of the denominator to eliminate the root term. Integration by trigonometric substitution we need to substitute the given variable equals to any trigonometric function. You should always choose which trigonometric function is suitable for solving the particular type of question. We should do all the calculations carefully and explicitly to avoid making errors.