# Homework Help: Trig Substitution with Integration

1. Dec 14, 2004

### 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.

2. Dec 14, 2004

### Hurkyl

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

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

3. Dec 14, 2004

### dextercioby

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

4. Dec 14, 2004

### Tom McCurdy

Would that just be

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

Last edited: Dec 14, 2004
5. Dec 14, 2004

### Hurkyl

Staff Emeritus
Yep (assuming you meant cos2 &theta;, and get your signs right)

6. Dec 14, 2004

### Tom McCurdy

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

7. Dec 14, 2004

### Pyrrhus

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

Hurkyl means

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

8. Dec 14, 2004

### 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.

9. Dec 14, 2004

### 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

10. Dec 14, 2004

### Hurkyl

Staff Emeritus
The same thing!

11. Dec 14, 2004

### Tom McCurdy

Thats the problem I have a test on all the material I miseed comming 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

12. Dec 14, 2004

### Tom McCurdy

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

13. Dec 14, 2004

### dextercioby

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

14. Dec 14, 2004

### Tom McCurdy

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

15. Dec 14, 2004

### Tom McCurdy

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

16. Dec 14, 2004

### cyby

That is a standard trig substitution, no?

17. Dec 14, 2004

### dextercioby

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.

Last edited: Dec 14, 2004
18. Dec 14, 2004

### 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???

Last edited: Dec 14, 2004
19. Dec 14, 2004

### 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)

20. Dec 14, 2004

### dextercioby

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.

21. Dec 14, 2004

### 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???

22. Dec 14, 2004

### dextercioby

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.

23. Dec 14, 2004

### dextercioby

$$u=\arg\cosh (\frac{x}{2})$$.

24. Dec 14, 2004

### 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???

25. Dec 15, 2004

### HallsofIvy

Thats a standard substitution. 1- sin2&theta;= cos2&theta; so $\sqrt{1- sin^2\theta}= cos \theta$. The substitution x= sin &theta; will simplify $\sqrt{1- x^2}$ to sin &theta;. Of course the substitution x= cos(&theta;) will also work.