Advanced Math/Proof for Integrals that result in inverse trigonometric functions
Expert: Steve Holleran - 2/25/2008
QuestionI 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.
AnswerHi 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