Trying an alternative to Trig Substitutions

I wanted to try an alternative method to the proverbial technique used in trig substitution. Is this method a dead-end or is there hope for it?

$$\int \frac{\sqrt{x^2-3}}{x} dx$$
Using trig substitution
$$c^2=a^2+b^2$$
$$a = \sqrt{c^2-b^2}$$
$$∴ c = x, b = \sqrt{3}$$
Assigning these values to a triangle, where a is adjacent to theta and b is opposite to theta we get..
$$\int \frac{\sqrt{x^2-3}}{x} dx = \int \frac{a}{x} dx = \int cos \theta dx$$

And I've hit a wall. I can't seem to be able to adequately convert dx to dθ.

DryRun
Gold Member
You could try the substitution: $$x=\sqrt 3 \cosh \theta$$

SammyS
Staff Emeritus
Homework Helper
Gold Member
I wanted to try an alternative method to the proverbial technique used in trig substitution. Is this method a dead-end or is there hope for it?

$$\int \frac{\sqrt{x^2-3}}{x} dx$$
Using trig substitution
$$c^2=a^2+b^2$$
$$a = \sqrt{c^2-b^2}$$
$$∴ c = x, b = \sqrt{3}$$
Assigning these values to a triangle, where a is adjacent to theta and b is opposite to theta we get..
$$\int \frac{\sqrt{x^2-3}}{x} dx = \int \frac{a}{x} dx = \int cos \theta dx$$

And I've hit a wall. I can't seem to be able to adequately convert dx to dθ.
Your substitution appears to be $\displaystyle \cos(\theta)=\frac{\sqrt{x^2-3}}{x}\ .$

In that case, how is dx related to dθ ?

You could try the substitution: $$x=\sqrt 3 \cosh \theta$$

I just did, it starts becoming ugly. And your still plagued by the dx so it hasn't made a difference.

Edit: I think I over-complicated things, I'm reviewing my method.

Edit: Either way, I feel a bit uncomfortable with the approach because I don't know how

$$x = \sqrt{3} cosh \theta$$

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Your substitution appears to be $\displaystyle \cos(\theta)=\frac{\sqrt{x^2-3}}{x}\ .$

In that case, how is dx related to dθ ?

Well, it doesn't seem that I can know until I can explicitly define the function in terms of x.

DryRun
Gold Member
It works out by the substitution that i suggested. What is the final answer given in your book/notes?

It works out by the substitution that i suggested. What is the final answer given in your book/notes?

$$\sqrt{3} \int (sec^2 \theta - 1) d\theta$$
$$\sqrt{3} (tan \theta - \theta)$$

dextercioby
Homework Helper
The substitution reccomended in post #2 should solve the problem.

The substitution reccomended in post #2 should solve the problem.

I can use any of the proverbial substitutions given in the book but that is besides the point. I was trying to do it in a different way and just need to know if my method is a dead end or not.

And at any rate, I'm not too familiar with the hyperbolic functions and how he derived that substitution equivalent.

HallsofIvy
Homework Helper
Why are the substitutions in your book "proverbial"? I don't recally any proverbs related to Calculus!

In any case, your method is not a "dead end" but it is exactly the trig substitution you apparently are trying to avoid. Taking x to be the hypotenuse of your right triangle and the "near side" to be $\sqrt{x^3- 3}$, the "opposite side" has length 3 so $sin(\theta)= 3/x$, $x= 3/sin(\theta)= 3csc(\theta)$. From that, $dx= 3 d(csc(\theta))= -3 csc(\theta)cot(\theta) d\theta$.

Why are the substitutions in your book "proverbial"? I don't recally any proverbs related to Calculus!

Proverbial as in widely known and common.

In any case, your method is not a "dead end" but it is exactly the trig substitution you apparently are trying to avoid. Taking x to be the hypotenuse of your right triangle and the "near side" to be $\sqrt{x^3- 3}$, the "opposite side" has length 3 so $sin(\theta)= 3/x$, $x= 3/sin(\theta)= 3csc(\theta)$. From that, $dx= 3 d(csc(\theta))= -3 csc(\theta)cot(\theta) d\theta$.

Hmm, I see now. ^.^ Thanks.