Finding Integrals Using Trigonometric Substitutions

Date: 10/2/95 at 8:12:18
From: roberto manzueta
Subject: What is the integral of 1/(x(1-x))^(1/2)?

Dr. Math,

I'm currently taking Calculus II.  I have not been able to 
integrate the above function with the methods I have learned so 
far.  The answer to  this problem is the arcsin [something] 
because I saw it at the back of my text (I do not remember that 
"something" right now).  Anyway, how can I integrate this function 
using only the derivative of arcsin or arctan?

Date: 10/20/95 at 14:11:18
From: Doctor Ken
Subject: Re: What is the integral of 1/(x(1-x))^(1/2)?

Hello!

Here's a hint about how to do your integral.  Notice that the  
denominator of the fraction looks like the square root of a 
quadratic polynomial.  This should make the buzzers in your brain 
go off and scream "TRIG SUBSTITUTION!!"  

Of course, it isn't in _exactly_ the right form for a trig 
substitution.  Remember that the derivative of Arcsin(t) is 
1/Sqrt{1 - t^2).  Well, your problem is how to turn the inside 
part, x(1-x) into something of the form 1-t^2.  

Hint: write x(1-x) as 1 - (1 - x(1-x)) and then complete the 
square on the 1 - x(1-x) part!

Hope this is enough for you to go on.  Good luck, and write back 
if you need more help!

-Doctor Ken,  The Geometry Forum