# Trigonometric substitution?

• Nov 7th 2006, 03:16 PM
FLTR
Trigonometric substitution?
Use trigonometric substitution to evaluate$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx$
• Nov 7th 2006, 03:32 PM
topsquark
Quote:

Originally Posted by FLTR
Use trigonometric substitution to evaluate$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx$

Let $\displaystyle x = tan \theta$. Then $\displaystyle dx = sec^2 \theta \, d \theta$

So
$\displaystyle \int \frac{1}{\sqrt{1+x^2}}dx = \int \frac{sec^2 theta}{\sqrt{1+tan^2 \theta}}d \theta$

= $\displaystyle \int \frac{sec^2 \theta}{\sqrt{sec^2 \theta}}d \theta$

= $\displaystyle \int \frac{sec^2 \theta}{sec \theta}d \theta$

= $\displaystyle \int sec \theta \, d \theta$

= $\displaystyle - ln \left | tan \theta - sec \theta \right |$

= $\displaystyle - ln \left | x - sec(atn(x)) \right |$

I leave the rest of the simplification to you.

-Dan
• Nov 7th 2006, 06:54 PM
ThePerfectHacker
First topsquark, I would like to mention that a composition of a trigonometric function and inverse function is some algebraic function. Furthermore, this simplicification is the inverse hyperbolic sine function.
• Nov 8th 2006, 04:16 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
First topsquark, I would like to mention that a composition of a trigonometric function and inverse function is some algebraic function. Furthermore, this simplicification is the inverse hyperbolic sine function.

I was aware of the first statement anyway.

However, I need to apologize, I am apparently off by at least one sign in my answer. Possibly because I set $\displaystyle \sqrt{xec^2 \theta} = sec \theta$ instead of the absolute value of such. (My usual forgetful mistake!)

-Dan
• Nov 8th 2006, 04:38 AM
topsquark
Quote:

Originally Posted by topsquark
I was aware of the first statement anyway.

However, I need to apologize, I am apparently off by at least one sign in my answer. Possibly because I set $\displaystyle \sqrt{xec^2 \theta} = sec \theta$ instead of the absolute value of such. (My usual forgetful mistake!)

-Dan

Nope! My solution was correct after all, just not in "standard form." So the answer is
$\displaystyle \int \frac{dx}{\sqrt{1+x^2}} = -ln \left | x - \sqrt{x^2+1} \right | = ln \left | x + \sqrt{x^2+1} \right |$

-Dan
• Nov 8th 2006, 07:27 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Nope! My solution was correct after all, just not in "standard form."

Your solution was not correct mathematically anyway. The manipulation of the differencials is stirctly prohibited. If you do it my way, by defining a composite function times its derivative then you can use the substitution rule, but that is slightly more difficult to get used to.
• Nov 8th 2006, 09:02 AM
Jameson
Quote:

Originally Posted by ThePerfectHacker
Your solution was not correct mathematically anyway. The manipulation of the differencials is stirctly prohibited. If you do it my way, by defining a composite function times its derivative then you can use the substitution rule, but that is slightly more difficult to get used to.

This is a fairly standard integral. What is incorrect about making the substitution $\displaystyle x=\sec(u)$?
• Nov 8th 2006, 09:28 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Your solution was not correct mathematically anyway. The manipulation of the differencials is stirctly prohibited. If you do it my way, by defining a composite function times its derivative then you can use the substitution rule, but that is slightly more difficult to get used to.

If you're complaining (again) about the $\displaystyle \frac{dx}{d \theta} = sec^2 \theta$ to $\displaystyle dx = sec^2 \theta d \theta$ transformation, no it is not strictly legitimate, but it has been proven to give correct results for any problem I've ever heard of. And I own 4 Introductory Calculus books that span some 20 years and all of them do integration substitutions using this notation.

I see no reason not to follow their example.

-Dan
• Nov 8th 2006, 11:01 AM
ThePerfectHacker
Quote:

Originally Posted by Jameson
This is a fairly standard integral. What is incorrect about making the substitution $\displaystyle u=\sec(x)$?

First you mean,
$\displaystyle \sec u=x$
Note when you differenciate you get the differencials,
$\displaystyle \tan^2 u du=dx$
But can you explain what those differencials mean!
They are not numbers!
Can you understand why it works? I cannot!

Thus, I use the strict form of the rule,
If $\displaystyle f(x)$ has an antiderivative on some open interval and $\displaystyle g(x)$ is differenciable on this interval then,
$\displaystyle \int (f\circ g)g' dx=F\circ g+C$
Where, $\displaystyle F$ is any function which satisfies the differencial equation $\displaystyle F'=f$ throughout the interval.

If you can formally prove to me how these differencials work then I will use them, but I never understood them and willing to bet most people do not understand them either (if any at all).

Another thing, which makes me really angry is the symbol $\displaystyle dx$ after the integral,I myself would rather drop it. In fact, my formal, differencial equations professor, does not write it. And my 12th Grade teacher also had a strong dislike to it, he used the composite function method (chain rule). Again I myself dislike it, greatly. I am sure if you open any analysis book they do not use this method, rather the correct one.

I believe these symbols are historical. They came from the time of Leibniz and remained until today. But the problem with Calculus (or "Method of Fluxions") in the 17 Century was that many mathemations disfavored it and opposed it (for example, Michelle' Rolle). They claimed it used "unsound reasoning", which in fact is true. Derivations using infitesimal arguments looked appealing but they were not based on mathematical derivations and thus were rejected by the mathematicians (but accepts by physcisits). Until the Cauchy and Weirestrauss became to build a foundation mathematicians did not approve Calculus.

I understand, topsquark, that this can produce the correct result but in strange cases this will probably not work. It is important for a mathematician to understand how something work rather than use it and know that it works. This is not alchemy.

I believe thee need a mathematicans apology.
• Nov 8th 2006, 12:56 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
First you mean,
$\displaystyle \sec u=x$
Note when you differenciate you get the differencials,
$\displaystyle \tan^2 u du=dx$
But can you explain what those differencials mean!
They are not numbers!
Can you understand why it works? I cannot!

Thus, I use the strict form of the rule,
If $\displaystyle f(x)$ has an antiderivative on some open interval and $\displaystyle g(x)$ is differenciable on this interval then,
$\displaystyle \int (f\circ g)g' dx=F\circ g+C$
Where, $\displaystyle F$ is any function which satisfies the differencial equation $\displaystyle F'=f$ throughout the interval.

If you can formally prove to me how these differencials work then I will use them, but I never understood them and willing to bet most people do not understand them either (if any at all).

Another thing, which makes me really angry is the symbol $\displaystyle dx$ after the integral,I myself would rather drop it. In fact, my formal, differencial equations professor, does not write it. And my 12th Grade teacher also had a strong dislike to it, he used the composite function method (chain rule). Again I myself dislike it, greatly. I am sure if you open any analysis book they do not use this method, rather the correct one.

I believe these symbols are historical. They came from the time of Leibniz and remained until today. But the problem with Calculus (or "Method of Fluxions") in the 17 Century was that many mathemations disfavored it and opposed it (for example, Michelle' Rolle). They claimed it used "unsound reasoning", which in fact is true. Derivations using infitesimal arguments looked appealing but they were not based on mathematical derivations and thus were rejected by the mathematicians (but accepts by physcisits). Until the Cauchy and Weirestrauss became to build a foundation mathematicians did not approve Calculus.

I understand, topsquark, that this can produce the correct result but in strange cases this will probably not work. It is important for a mathematician to understand how something work rather than use it and know that it works. This is not alchemy.

I believe thee need a mathematicans apology.

(Shrugs) Suit yourself, I'm not about to get into an argument over it. I've never taken a full-fledged Analysis class (where you get into the picky details of why Calculus works) though I would like to some day. And whereas I know you are correct about the treatment of differential elements, this is the way that (as far as I know) practically all Introductory level Calculus classes approach the topic of substitution of variables in integrals. I am willing to learn new (to me) and more precise techniques, but since the "language" of most of the students on the forum is likely to be the same as mine in this matter I will likely continue to speak using it.

Just giving you apologies in advance for my "informal" methods! :)

-Dan