# Definite integral with square on bottom

## Homework Statement

$\int_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}}dt$

## The Attempt at a Solution

Ok, this is an odd problem I'm working from the book. The book says the answer is $\pi$.

First I tried getting rid of the fraction:

$\frac{6}{\sqrt{1-t^2}}dt$ = $\frac{6}{(1-t^{2})^{1/2}}$ = $6(1-t^{2})^{-1/2}$

I think I may be getting this next step wrong, but next, I did this:

= $6(1^{1/2}-t^{3/2})$ = $6^{1/2}-6t^{3/2}$

Which I integrated to:

$6t^{1/2} - \frac{12}{5}t^{5/2}$

I can't seem to plug in my numbers and get $\pi$ though.

Where did I mess up so far?

Thanks for taking the time to read through my poor attempt at solving this.

I think I may be getting this next step wrong, but next, I did this:

= $6(1^{1/2}-t^{3/2})$ = $6^{1/2}-6t^{3/2}$

This step is wrong. I can't even see what you did there. Basically, you cannot get rid of the square root by splitting it over a sum, something like that is not allowed.

Try to do a trigonometric substitution. A subsitution like t=sin(u)

$6(1-t^{2})^{-1/2}$

I tried to take the -1/2 exponent and add it to the 1-t^2 which gave me this:

= $6(1^{1/2}-t^{3/2})$ = $6^{1/2}-6t^{3/2}$

I figured I was doing it wrong, but my problem is that I'm not quite sure how to integrate $6(1-t^{2})^{-1/2}$

I'm not really sure how to change it into another form either.

This step is wrong. I can't even see what you did there. Basically, you cannot get rid of the square root by splitting it over a sum, something like that is not allowed.

Try to do a trigonometric substitution. A subsitution like t=sin(u)

I'm not sure I understand how substituting in sin would help.

I'm not sure I understand how substituting in sin would help.

It would get rid of the square root...

What happens if you substitute in sin(u)??

factor out the 6, then the integral become arcsin(t)

I don't think we have covered anything on arcsin yet. I'm only in Calculus 2. If I should know that at this level, I guess one of my teachers let me down.

Well I had AP Calculus last year and we covered that, but I'm in Calculus 2 this year and we haven't gotten that far yet in class but it'll come up later. Same thing with trig substitutions as someone previously stated.

SammyS
Staff Emeritus
Homework Helper
Gold Member

Do as micromass suggested. Let t=sin(u) , then dt = cos(u) du .

Yes, that's equivalent to letting u = arcsin(t), but you won't need that fact until plugging-in the limits of integration, and at that point it should pose no problem.

BTW, 1 - sin2(θ) = cos2(θ) .

Nope. We haven't had substitutions yet either. It looks like that is the next chapter we will cover if we stick to the order of the book.

Nope. We haven't had substitutions yet either. It looks like that is the next chapter we will cover if we stick to the order of the book.

OK, so you can't use substitution??

Well, do you have a list of standard integrals, then??

After looking through the book, I found a table of integrals that has the inverse sin function. I don't seem to have the inverse ones in my notes. I don't know how I missed that. I got it figured out now. Thanks for the help.