Homework Help: Trig. substitution

1. Feb 7, 2005

Consider the definite integral $$\int \frac{(x^3)}{(sqrt(3x^2-1))}$$

can someone help me find the appropriate subsitution?

i know that i will need this subsitution:

$$sqrt(x^2-a^2)$$ is equal to
$$x=a*sec(theta)$$

well... i have to make 3x^2 look like x^2 somehow.

i tried using u-du sub, but i cant really find the right subsitution.

can someone give me a hand?

2. Feb 7, 2005

dextercioby

The hard way:Do you handle calculus with hyperbolic functions...?

The easy way:Try part integration.

Daniel.

3. Feb 7, 2005

MathStudent

There is also a very messy way by using

$$x= \frac{1}{\sqrt{3}} \sec \theta$$

but it will get the job done.

4. Feb 8, 2005

perfect! thank you

btw, how did you get $$x= \frac{1}{\sqrt{3}} \sec \theta$$ ? i have my thoughts on how you got it, but i would like to make sure.

5. Feb 8, 2005

MathStudent

pretty standard, when you have

$$\sqrt{bx^2 - a^2}$$

factor out the b to give you the recognizable $x^2 - c^2$ form so that you have

$$\sqrt{b(x^2 - \frac{a^2}{b})}$$

and then make the substitution

$$x = \frac{a}{\sqrt{b}}\sec \theta$$

Last edited: Feb 8, 2005
6. Feb 8, 2005

dextercioby

Pretty standard with cosine hyperbolic too... :tongue:

Daniel.

7. Feb 8, 2005

MathStudent

It seems like most people, when given the chance avoid the hyperbolics like the plague, I don't even think they're taught at some shools.

8. Feb 8, 2005

dextercioby

I'd never use secant & cosecant in any of my formulas...Any neither their hyperbolic or elliptical counterparts...Actually,i've always screwed them up.
secant-------->sinus
cosecant-------cosinus

would have been much more helpful for my slow brain.

Daniel.