How Do Partial Fractions Relate to Trigonometric Substitution?

[askor]

Do anyone know how to find $1$, $2x - 5$, and $2\sqrt{x^2 - 5x + 6}$ in the triangle? (please see attached image)

Also, how do you find $(x - 5/2)^2 - (1/2)^2$?

[Moderator's note: Moved from a technical forum and thus no template.]

 

[PeroK]

Do anyone know how to find $1$, $2x - 5$, and $2\sqrt{x^2 - 5x + 6}$ in the triangle? (please see attached image)

Also, how do you find $(x - 5/2)^2 - (1/2)^2$?

The first one is called Pythagoras's theorem; and the second is completing the square for a quadratic expression.

 

[askor]

The first one is called Pythagoras's theorem; and the second is completing the square for a quadratic expression.

Yes I know how to use Pythagoras' theorem and completing the square. But where are they come from? I don't understand. Anyone mind to explain as detail as possible?

 

[PeroK]

Yes I know the first is Phytagoras theorem and second is completing square. But where are they come from? I don't understand. Anyone mind to explain as detail as possible?

What do you mean "where are they come from"? You have a quadartic, you complete the square and you draw the corresponding right-angle triangle.

Where did $x^2 - 5x + 6$ come from?

 

[askor]

What do you mean "where are they come from"? You have a quadartic, you complete the square and you draw the corresponding right-angle triangle.

Where did $x^2 - 5x + 6$ come from?

The $x^2 - 5x + 6$ is from the problem: $\int \frac{1}{x^2 - 5x + 6} dx$

How do you get $1$, $2x - 5$, and $2 \sqrt{x^2 - 5x + 6}$ in the triangle?

 

[PeroK]

The $x^2 - 5x + 6$ is from the problem: $\int \frac{1}{x^2 - 5x + 6} dx$

How do you get $1$, $2x - 5$, and $2 \sqrt{x^2 - 5x + 6}$ in the triangle?

Pythagoras theorem. Try it!

 

[askor]

Why the $1$ is in adjacent side, why $2x - 5$ is in hypotenuse side and why $2 \sqrt{x^2 - 5x + 6}$ is in opposite side?

 

[PeroK]

Why the $1$ is in adjacent side, why $2x - 5$ is in hypotenuse side and why $2 \sqrt{x^2 - 5x + 6}$ is in opposite side?

Okay, you've clearly got a mental block about this. You start with:

$x^2 - 5x + 6 = (x- 5/2)^2 - (1/2)^2$

Hence:

$x^2 - 5x + 6 + (1/2)^2 = (x- 5/2)^2$

You multiply by $4$ to get rid of the fractions:

$4(x^2 - 5x + 6) + 1^2 = (2x- 5)^2$

You use Pythagoras to justify drawing a right triangle with $(2x-5)$ as the hypothenuse and $1$ and $2\sqrt{x^2 - 5x + 6}$ as the other two sides.

Simple!

 

[askor]

Okay, you've clearly got a mental block about this. You start with:

$x^2 - 5x + 6 = (x- 5/2)^2 - (1/2)^2$

Hence:

$x^2 - 5x + 6 + (1/2)^2 = (x- 5/2)^2$

You multiply by $4$ to get rid of the fractions:

$4(x^2 - 5x + 6) + 1^2 = (2x- 5)^2$

You use Pythagoras to justify drawing a right triangle with $(2x-5)$ as the hypothenuse and $1$ and $2\sqrt{x^2 - 5x + 6}$ as the other two sides.

Simple!

This is the answer I want. Why don't you answer my question like this earlier?

 

[PeroK]

This is the answer I want. Why don't you answer my question like this earlier?

Because this is elementary mathematics and you are studying the more advanced mathematics of trig integral substitutions. The ethos of this site is to get you to think for yourself. In this case, you may have thought something complicated was going on. My first posts hinted that there was nothing complicated going on and you just needed to do some basic algebra. Which, as it turned out, you were unable to do for yourself.

 

[askor]

More question, why do the form change from $x^2 - 5x + 6$ to $(x - 5/2)^2 - (1/2)^2$ ?

 

[PeroK]

More question, why do change from $x^2 - 5x + 6$ to $(x - 5/2)^2 - (1/2)^2$

Why not?

You will do that a lot in integration when you have a quadratic on the denominator. The idea is to get rid of the linear term. With the substitution $u = x - 5/2$ you have a denominator that looks like $u^2 - a^2$, where $a$ is a constant. These are often standard integrals, which can be solved by a trig or hyperbolic trig substitution. Or, looked up in tables.

 

[askor]

Why not?

You will do that a lot in integration when you have a quadratic on the denominator. The idea is to get rid of the linear term. With the substitution $u = x - 5/2$ you have a denominator that looks like $u^2 - a^2$, where $a$ is a constant. These are often standard integrals, which can be solved by a trig or hyperbolic trig substitution. Or, looked up in tables.

I only know trigonometric substitution from my calculus textbook but not hyperbolic trig substitution, I can't find it in my calculus textbook. In what book the hyperbolic substitution is teached?

 

[PeroK]

I only know trigonometric substitution from my calculus textbook but not hyperbolic trig substitution, I can't find it in my calculus textbook. In what book the hyperbolic substitution is teached?

It's in the book of Google!

 

[askor]

It's in the book of Google!

Come on, be serious. What book is it?

Back to the topic. What is the relationship of partial fractions with trigonometric substitution?

 

[PeroK]

Come on, be serious. What book is it?

Back to the topic. What is the relationship of partial fractions with trigonometric substitution?

That's what your book is trying to show you.

 

[SammyS]

Do anyone know how to find $1$, $2x - 5$, and $2\sqrt{x^2 - 5x + 6}$ in the triangle? (please see attached image)

Also, how do you find $(x - 5/2)^2 - (1/2)^2$?

When posting images, please, post the full sized version in addition to the thumbnail.

 

[SammyS]

...

Back to the topic. What is the relationship of partial fractions with trigonometric substitution?

PeroK did answer this in a short statement. Here's a longer explanation.

The attachment included in the OP gives an example of an integral which can be evaluated using either trig substitution or by decomposing the integrand using the method that partial fractions. For this example partial fractions gives the result more directly. However, the excerpt you show, only gives the result using trig substitution. I assume that your textbook also evaluates the integral using partial fractions.

Use partial fraction decomposition to show that $\dfrac {1}{x-2}- \dfrac {1}{x-3} = \dfrac {1}{ x^2-5x+6}$

If you remain curious about Hyperbolic Substitution, Look at the Wikipedia page on Trig Substitution to see a section on Hyperbolic Substitution.