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Question: How do you integrate\[\int{\dfrac{1}{{{x}^{2}}\sqrt{9-{{x}^{2}}}}dx}\] using trigonometric substitut...

How do you integrate∫1x29−x2dx\int{\dfrac{1}{{{x}^{2}}\sqrt{9-{{x}^{2}}}}dx} using trigonometric substitution?

Explanation

Solution

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,
⇒∫1x29−x2dx\Rightarrow \int{\dfrac{1}{{{x}^{2}}\sqrt{9-{{x}^{2}}}}dx}
Let the given integral be I.
⇒I=∫1x29−x2dx\Rightarrow I=\int{\dfrac{1}{{{x}^{2}}\sqrt{9-{{x}^{2}}}}dx}
Substitutex=3sin⁡u then dx=3cos⁡udux=3\sin u\ then\ dx=3\cos udu,
⇒I=∫1(3sin⁡u)29−9sin⁡2u3cos⁡udu\Rightarrow I=\int{\dfrac{1}{{{\left( 3\sin u \right)}^{2}}\sqrt{9-9{{\sin }^{2}}u}}3\cos udu}
Simplifying the above, we get
⇒I=∫1(9sin⁡2u)31−sin⁡2u3cos⁡udu\Rightarrow I=\int{\dfrac{1}{\left( 9{{\sin }^{2}}u \right)3\sqrt{1-{{\sin }^{2}}u}}3\cos udu}
By using trigonometric identity, i.e. 1−sin⁡2x=cos⁡2x1-{{\sin }^{2}}x={{\cos }^{2}}x
We get,
⇒I=∫1(9sin⁡2u)3cos⁡2u3cos⁡udu\Rightarrow I=\int{\dfrac{1}{\left( 9{{\sin }^{2}}u \right)3\sqrt{{{\cos }^{2}}u}}3\cos udu}
Cancelling out 3cos (u), we get
⇒I=∫1(9sin⁡2u)du=19∫1sin⁡2udu=19∫cos⁡ec2udu\Rightarrow I=\int{\dfrac{1}{\left( 9{{\sin }^{2}}u \right)}du}=\dfrac{1}{9}\int{\dfrac{1}{{{\sin }^{2}}u}du}=\dfrac{1}{9}\int{\cos e{{c}^{2}}udu}
The resultant integral is a standard integral.
Thus,
⇒I=−19cot⁡u+C\Rightarrow I=-\dfrac{1}{9}\cot u+C
Since, as we know that
By using Pythagoras theorem,
sin⁡u=x3,cos⁡u=9−x23\sin u=\dfrac{x}{3},\cos u=\dfrac{\sqrt{9-{{x}^{2}}}}{3}
Thus,
cot⁡u=cos⁡usin⁡u=9−x23x3=9−x2x\cot u=\dfrac{\cos u}{\sin u}=\dfrac{\dfrac{\sqrt{9-{{x}^{2}}}}{3}}{\dfrac{x}{3}}=\dfrac{\sqrt{9-{{x}^{2}}}}{x}
Putting the value of cot (u) in the above integral, we get
⇒I=−19cot⁡u+C=−19(9−x2x)+C\Rightarrow I=-\dfrac{1}{9}\cot u+C=-\dfrac{1}{9}\left( \dfrac{\sqrt{9-{{x}^{2}}}}{x} \right)+C

Therefore,
⇒∫1x29−x2dx=−19(9−x2x)+C\Rightarrow \int{\dfrac{1}{{{x}^{2}}\sqrt{9-{{x}^{2}}}}dx}=-\dfrac{1}{9}\left( \dfrac{\sqrt{9-{{x}^{2}}}}{x} \right)+C
Hence, it is the required answer.

Note: We should remember the property or the formulas of integration, this would make it easier to solve the question. Students need to remember that to eliminate the root term we need to multiply the numerator and denominator by the conjugate of the denominator. 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. You should always remember all the methods for integration so that we can easily choose which method is suitable for solving the particular type of question.