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Trigonometric substitution


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#1 DylsexicChciken

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Posted 12 September 2014 - 03:33 AM

In solving an integral of sqrt(9-x2), we substitute x=3sin(t) to get an integral of 3cos(t). Why does this work? In the original function, x can take on any value(if you include imaginary numbers), but the new function is equivalent to the original one even though sin(t) takes values only between its amplitudes.

 

P.S. I remember a guide to writing math in this forum. Anyone have the link?


Edited by DylsexicChciken, 12 September 2014 - 03:35 AM.

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#2 John

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Posted 12 September 2014 - 05:48 AM

It works the same as any other substitution. You're introducing a new variable t and making x a function of t, with the goal of working with a simpler integrand.

The function is probably intended to be a real-valued function, as introducing complex numbers leads into various shenanigans generally beyond the scope of introductory calculus. Thus x in this case can only take values in the interval [-3, 3]. Notice that since x = 3sint, we preserve this range of values for x, i.e. since sint ranges from -1 to 1, 3sint ranges from -3 to 3.

If you'd like to see the reasoning behind why integration by substitution works, then check out this ProofWiki article.

Dave's quick guide to using LaTeX on the forum is the first stickied thread in the main Mathematics forum, but here is a link anyway.


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#3 Amad27

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Posted 14 September 2014 - 09:53 AM

Hello,

 

If you draw a right triangle, you will clearly see the relation between what you had. 

 

If you have an angle suppose t then you can relate it to x to simplify and integrate.

 

Advice: Draw a right triangle and then substitute, you'll see how it works (using the Pythagorean Theorem, ofcourse). 


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#4 deesuwalka

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Posted 13 October 2016 - 09:36 AM

We substitute  \sqrt{9-x^2}   by   x= 3sin\theta   because a trigonometric substitution</a> helps us to get a perfect square under the radical sign. This simplifies the integrand function.http://www.actucatio...etric-functions

 

Now, we can simplified it easily,

 

 \int\sqrt{9-x^2}\;\;\implies\int\sqrt{3^2-x^2}

 

 

  \int\sqrt{3^2-x^2}\;\implies\sqrt{3^2-(3sin\theta)^2}

 

 

 =\sqrt{9-9sin^2\theta}

 

 

 

 \sqrt{9-9sin^2\theta}\;\;\implies\sqrt{9(1-sin^2\theta)}\;\;\implies\sqrt{9cos^2\theta}\;=\;\int 3cos\theta

 

I hope it' ll help.


Edited by deesuwalka, 13 October 2016 - 10:44 AM.

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#5 HallsofIvy

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Posted 14 October 2016 - 11:39 AM

  You don't "include imaginary numbers"!  This integral is over the real numbers.


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