May i know how to integrate [1 / (4 + x^2)^1/2] using trig?

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Homework Help Overview

The discussion revolves around the integration of the function [1 / (4 + x^2)^(1/2)] using trigonometric substitution, a topic within calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks guidance on how to approach the integration using trigonometric substitution. Some participants suggest specific substitutions and discuss relevant trigonometric identities that may apply. Others inquire about general strategies for selecting appropriate identities in different integration problems.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for substitutions and discussing the applicability of trigonometric identities. There is a focus on understanding the reasoning behind choosing specific identities for various forms of integrals.

Contextual Notes

There is an emphasis on the need for understanding which trigonometric identities are useful in different contexts, particularly for integration involving radicals. The discussion reflects a learning environment where participants are encouraged to explore and clarify their understanding of trigonometric substitutions.

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may i know how to integrate [1 / (4 + x^2)^1/2] using trigo substitution??

pls help...
 
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Well, you have a few trig identities, see which one is useful.
Here, you could use the fact that [itex]\sec ^2 x = 1 + \tan ^2 x[/itex].

So try the substitution [itex]x = 2\tan y \Leftrightarrow dx = \frac{2}{{\cos ^2 y}}dy[/itex]
 
oh,i got it...can u pls tell me how do you know which frigo identities to choose when u you see the question in general??(not the question above)

pls...
 
For trigonometric substitutions, there are two fundamental identities which are often used:

[tex]\begin{array}{l}<br /> \cos ^2 x + \sin ^2 x = 1 \Leftrightarrow \cos ^2 x = 1 - \sin ^2 x \\ <br /> \sec ^2 x = 1 + \tan ^2 x \Leftrightarrow \tan ^2 x = \sec ^2 x - 1 \\ <br /> \end{array}[/tex]

The first one can be used for radical expressions of the form [itex]\sqrt {a^2 - x^2 }[/itex] where you then choose the substitution [itex]x = a\sin y[/itex].
The second one can be used for two types: yours, which was of the form [itex]\sqrt {x^2 + a^2 }[/itex] (you then do [itex]x = a\tan y[/itex]) or those of the form [itex]\sqrt {x^2 - a^2 }[/itex] (then it's [itex]x = a\sec y[/itex]).
 

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