Trig Substitution with Integration

[Tom McCurdy]

How would you go about solving
$$\int \frac{\sqrt{1-x^2}}{x^2}$$ ?

I have tried a few things... drawing out triangles... etc but can't seem to get it... I am kind of behind in math because I was gone for awhile because of being sick and presentations.

 

[Hurkyl]

Ok, let's take a step back...

Can you do $\int \sqrt{1-x^2} \, dx$?

 

[dextercioby]

How would you go about solving
$$\int \frac{\sqrt{1-x^2}}{x^2}$$ ?

I have tried a few things... drawing out triangles... etc but can't seem to get it... I am kind of behind in math because I was gone for awhile because of being sick and presentations.

Try the substitution $x\rightarrow \sin u$.And then part integration.
Daniel

 

[Tom McCurdy]

Would that just be

$$x=sin\theta$$
$$\sqrt{1-sin^2\theta$$
$$sin^2=1-cos^2\theta$$
$$\sqrt{1-(1-cos^2\theta)$$
$$\int \sqrt{-cos^2\theta$$

 

[Hurkyl]

Yep (assuming you meant cos² θ, and get your signs right)

 

[Tom McCurdy]

hmm alright I am still lost even on your back up step

 

[Pyrrhus]

Tom, again i emphatize on the important of the differential... do not forget about putting them on your integrals.

Hurkyl means

$$\sqrt{1 - \sin^2} = \sqrt{\cos^2}$$

 

[PICsmith]

As long as you're not in the middle of a test or something you can look it up in integral tables and see what answer they got and from that determine what methods they used. For example the solution to this integral is in a form that looks like it was done by parts, plus it has an inverse sine in it, which hints at trig substitutions as was mentioned by dextercioby.

 

[Tom McCurdy]

wow... i missed that competely i need more sleep... so what would I do to for the orignial problem with x^2 in the denominator

 

[Hurkyl]

The same thing!

 

[Tom McCurdy]

As long as you're not in the middle of a test or something you can look it up in integral tables and see what answer they got and from that determine what methods they used. For example the solution to this integral is in a form that looks like it was done by parts, plus it has an inverse sine in it, which hints at trig substitutions as was mentioned by dextercioby.

Thats the problem I have a test on all the material I miseed coming up on thursday... and I need to make sure I get a good grade in the class if I want any chance of getting accepted after getting defred from MIT

 

[Tom McCurdy]

So would it just come out to be $$\int \frac{\sqrt{cos^2\theta}}{sin^2\theta}$$ ?

 

[dextercioby]

So would it just come out to be $$\int \frac{\sqrt{cos^2\theta}}{sin^2\theta}$$ ?

Not exactly.U need to transform "dx" as well.That will give another "cosine".
In should be
$$\int \frac{\cos^{2}\theta}{\sin^{2}\theta}d\theta$$.

 

[Tom McCurdy]

oh yeah I forgot about that... so it becomes the $$\int tan^2\theta$$

 

[Tom McCurdy]

My question is how did you decide to make $x=sinu$

 

[cyby]

That is a standard trig substitution, no?

 

[dextercioby]

My question is how did you decide to make $$x=sinu$$

1.First of all it's $$\arctan^{2}\theta$$.
2.Experience at doing integrals? Actually it was the expression under the radical that led to the natural substitution "sine"/"cosine",just because:
$$1-\sin^{2}\theta =\cos^{2}\theta$$ and another one similar.

Daniel.

 

[Tom McCurdy]

Alright I am going to assume $x=4sinu$ for

$$\int \frac{x^3}{\sqrt{x^2-4}}dx$$

Therefore
$$dx=cos\theta$$

$$\int \frac{4sin\theta^3}{\sqrt{4sin\theta^2-4}}*cos\theta$$

$$4 \int \frac{sin\theta^3}{\sqrt{sin^2\theta-1}} *cos\theta$$

$$4 \int \frac{sin\theta^3}{\sqrt{cos^2}}*cos\theta$$

$$4\int sin\theta^3 d\theta$$ ?


Did i do it right?

 

[Parth Dave]

it depends on what form it is in, if you have:
(a^2 - x^2) : x = asin(theta)
(a^2 + x^2) : x = atan(theta)
(x^2 - a^2) : x = asec(theta)

 

[dextercioby]

Alright I am going to assume $x=4sinu$ for

$$\int \frac{x^3}{\sqrt{x^2-4}}dx$$

Therefore
$$dx=cos\theta$$
$$\int \frac{4sin\theta^3}{\sqrt{4sin\theta^2-4}}*cos\theta$$

work in progress...

Hold on,what u posted is wrong:Û don't need 4,but 2 (check the denominator)
$$dx=2 \cos\theta d\theta$$
$$x^{3}=8\sin^{3}\theta$$

Daniel.

 

[Tom McCurdy]

Alright I am going to assume $x=2sinu$ for

$$\int \frac{x^3}{\sqrt{x^2-4}}dx$$

Therefore
$$dx=cos\theta$$

$$\int \frac{8sin\theta^3}{\sqrt{4sin\theta^2-4}}*cos\theta$$

$$2 \int \frac{sin\theta^3}{\sqrt{sin^2\theta-1}} *cos\theta$$

$$2 \int \frac{sin\theta^3}{\sqrt{cos^2}}*cos\theta$$

$$2\int sin\theta^3 d\theta$$ ?


Did i do it right?

 

[dextercioby]

Did I do it right?

Obviously not!I'm sorry for giving you a wrong substitution:trigonometric hyperboilic functions (instead of the circular ones) should do the trick.

Take a good look here:
$$\int \frac{x^{3}}{\sqrt{x^{2}-4}} dx =...?$$
$$x\rightarrow 2\cosh u;dx\rightarrow 2 \sinh u du$$
The integral becomes:
$$\int \frac{8\cosh^{3} u}{\sqrt{4\cosh^{2}u -4}} 2\sinh u du$$
,which can be put in a form:
$$8\int \cosh^{3} u du$$
,where u have made use of the fundamental formula of the hiperbolic trigonometry:
$$\cosh^{2} u -\sinh ^{2} u =1$$.
Use the same formulla to evaluate that integral.

U should be gettin'
$$8(\sinh u +\frac{\sinh^{3} u}{3})$$.

Daniel.

 

[dextercioby]

And then,of course,u have to return to the original variable:
$$u=\arg\cosh (\frac{x}{2})$$.

 

[dextercioby]

If u don't like hyperbolic trigonometric functions,u can do a simple part integral.And the result should be the same.

Daniel.

PS.Do you see that part integral?

 

[HallsofIvy]

Thats a standard substitution. 1- sin²θ= cos²θ so $\sqrt{1- sin^2\theta}= cos \theta$. The substitution x= sin θ will simplify $\sqrt{1- x^2}$ to sin θ. Of course the substitution x= cos(θ) will also work.