Trying an alternative to Trig Substitutions

[Nano-Passion]

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]

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

 

[SammyS]

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 ?

 

[Nano-Passion]

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

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

 

[Nano-Passion]

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]

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

 

[Nano-Passion]

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]

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

 

[Nano-Passion]

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]

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, $dx= 3 d(csc(\theta))= -3 csc(\theta)cot(\theta) d\theta$.

 

[Nano-Passion]

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.