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Question: How do you integrate \[\int{\dfrac{1}{u\sqrt{5-{{u}^{2}}}}}\] using trigonometric substitution?...

How do you integrate ∫1u5−u2\int{\dfrac{1}{u\sqrt{5-{{u}^{2}}}}} 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,
⇒∫1u5−u2\Rightarrow \int{\dfrac{1}{u\sqrt{5-{{u}^{2}}}}}
Let the given integral be I.
⇒I=∫1u5−u2\Rightarrow I=\int{\dfrac{1}{u\sqrt{5-{{u}^{2}}}}}
Substituteu=5sin⁡θ then du=5cos⁡θdθu=\sqrt{5}\sin \theta \ then\ du=\sqrt{5}\cos \theta d\theta ,
⇒I=∫1(5sin⁡θ)5−5sin⁡2θ5cos⁡θdθ\Rightarrow I=\int{\dfrac{1}{\left( \sqrt{5}\sin \theta \right)\sqrt{5-5{{\sin }^{2}}\theta }}\sqrt{5}\cos \theta d\theta }
Simplify the above expression, we get
⇒I=∫1(5sin⁡θ)5(1−sin⁡2θ)5cos⁡θd\Rightarrow I=\int{\dfrac{1}{\left( \sqrt{5}\sin \theta \right)\sqrt{5\left( 1-{{\sin }^{2}}\theta \right)}}\sqrt{5}\cos \theta d}
By using trigonometric identity, i.e. 1−sin⁡2x=cos⁡2x1-{{\sin }^{2}}x={{\cos }^{2}}x
We get,
⇒I=∫1(5sin⁡θ)5cos⁡2θ5cos⁡θdθ\Rightarrow I=\int{\dfrac{1}{\left( \sqrt{5}\sin \theta \right)\sqrt{5{{\cos }^{2}}\theta }}\sqrt{5}\cos \theta d\theta }
⇒I=∫1(5sin⁡θ)5cos⁡θ5cos⁡θdθ\Rightarrow I=\int{\dfrac{1}{\left( \sqrt{5}\sin \theta \right)\sqrt{5}\cos \theta }\sqrt{5}\cos \theta d\theta }
Cancelling out 5cos⁡θ\sqrt{5}\cos \theta , we get
⇒I=∫1(5sin⁡θ)dθ\Rightarrow I=\int{\dfrac{1}{\left( \sqrt{5}\sin \theta \right)}d\theta }
⇒I=∫15cos⁡ecθdθ\Rightarrow I=\int{\dfrac{1}{\sqrt{5}}\cos ec\theta d\theta }
Taking the constant part out of the integral, we get
⇒I=15∫cos⁡ecθdθ\Rightarrow I=\dfrac{1}{\sqrt{5}}\int{\cos ec\theta d\theta }
The resultant integral is a standard integral.
As, we know that ∫cos⁡ex(x)dx=−ln⁡∣cos⁡ec(x)+cot⁡(x)∣\int{\cos ex\left( x \right)dx=-\ln \left| \cos ec\left( x \right)+\cot \left( x \right) \right|}
⇒I=−15ln⁡∣cos⁡ec(x)+cot⁡(x)∣\Rightarrow I=-\dfrac{1}{\sqrt{5}}\ln \left| \cos ec\left( x \right)+\cot \left( x \right) \right|
Since, as we know that
Using our above substitution,
sin⁡θ=u5, therefore cos⁡ecθ=1sin⁡θ=5u\sin \theta =\dfrac{u}{\sqrt{5}},\ therefore\ \cos ec\theta =\dfrac{1}{\sin \theta }=\dfrac{\sqrt{5}}{u}
Thus,
cot⁡θ=5−u2u\cot \theta =\dfrac{\sqrt{5-{{u}^{2}}}}{u}
Putting the value of cot⁡θ and cos⁡ecθ\cot \theta \ and\ \cos ec\theta in the above integral, we get
⇒I=−15ln⁡∣5+5−u2u∣+C\Rightarrow I=-\dfrac{1}{\sqrt{5}}\ln \left| \dfrac{\sqrt{5}+\sqrt{5-{{u}^{2}}}}{u} \right|+C

Therefore,
⇒∫1u5−u2=−15ln⁡∣5+5−u2u∣+C\Rightarrow \int{\dfrac{1}{u\sqrt{5-{{u}^{2}}}}}=-\dfrac{1}{\sqrt{5}}\ln \left| \dfrac{\sqrt{5}+\sqrt{5-{{u}^{2}}}}{u} \right|+C
Hence, it is the required answer.

Note: In mathematics, trigonometric substitution is the substitution of trigonometric functions from other expressions. It is a technique for evaluating integrals with the help of calculus concepts. While solving the above question be careful with the integration part. Do remember the substitution method used here for the future use. To solve these types of questions, we should have the knowledge of trigonometric identities. Do not forget to reverse the substitution or to undo the substitution and after integration add the constant of integration in the result.