trig. substitution

[ProBasket]

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?

[dextercioby]

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

The easy way:Try part integration.

Daniel.

[MathStudent]

There is also a very messy way by using

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

but it will get the job done.

[ProBasket]

There is also a very messy way by using

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

but it will get the job done.

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.

[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

[dextercioby]

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

Daniel.

[MathStudent]

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

Daniel.
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. :smile:

[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. :rolleyes:

Daniel.