You are here:

Advanced Math/Proof for Integrals that result in inverse trigonometric functions

Advertisement


Question
I need to write proofs for the following:
1. The integration of 1/squareroot[a^2 - x^2]dx when a is a constant.
2. The integration of 1/[a^2 + x^2]dx when a is a constant.
3. The integration of 1/[(u)squareroot[x^2 - a^2]] when a is a constant.
I know what the integration of each is, but I don't know how to begin the proof. I'm assuming each proof will be similar, so if you could just help me by beginning one of the proofs I would appreciate it.

Answer
Hi Allyson,

Well, I'm not very good at proof, but here's what I would do:

1.  INT [ 1/sqrt(a^2 - x^2) dx]  

   In the integrand, factor a^2 out inside the square root:

   sqrt(a^2 - x^2) = sqrt( a^2(1 - (x^2/a^2))

   Now take the a^2 out as |a|:

  = 1/|a| * 1/sqrt(1 - (x/a)^2) * dx

Then u = x/a and du = 1/a, so multiply outside the INT by a and inside by 1/a, and you have the form:

      a * INT[1/|a| * 1/sqrt(1 - u^2) * 1/a * du

Take the 1/|a| out of the INT also, and the form inside is an inverse sine antiderivative:

    =  a * 1/|a| * Arcsin(u) + C

    = +/- Arcsin(x/a) + C

The +/- depends on a, since a * 1/|a| can be +1 or -1.

That's the best I can do here.

Steve

Advanced Math

All Answers


Answers by Expert:


Ask Experts

Volunteer


Steve Holleran

Expertise

I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience

33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

©2012 About.com, a part of The New York Times Company. All rights reserved.