Why does the trigonometric substitution x = a*sec(u) hold for any x, given a is a constant?

Question

I know this is a basic substitution question but I haven't been able to figure it out. I often use trigonometric substitutions for integrands involving expressions such as $a^2-x^2$ and others. I have learnt to solve these by using the substitution, $x=a\,sin(u)$ and then the identity $cos^2(u)=1-sin^2(u)$.

However, I can't work out why I can use this substitution. How can the substitution $a\,sin(u)$ represent any possible real value of $x$, if $a$ is some given constant. For example, given $3^2-x^2$ I will use the substitution $x=3 \, sin(u)$ to represent the real number $x$. But, surely this means the number $x$ is limited to be $x \in [-a,a]$ rather than to be $x \in (- \infty, \infty)$?

I'm sure this is a very basic question and I must have missed something very obuvious. Any help would be much appreciated! Many thanks :)

I am confused for integrating $x^2$ why would some one do substition is it $1/(..)$ — Archis Welankar, May 07 '16 at 10:27
Yes, the term $a^2-x^2$ does appear under a square root. And, sorry, my title to the question is wrong it should say $x=a sin(u)$ rather than $x=a sec(u)$. Thanks :) — Henry_, May 07 '16 at 15:18

score 2 · Answer 1 · answered May 07 '16 at 15:23

With help from Karan Singh I have now worked it out! The expression $a^2-x^2$ always appears under a square root sign. Thus, $x^2 \leq a^2$ for the square root to produce any real answers! In other words, $x$ must be within the range $[-a,a]$ and so the substitution $x=a\,sin(u)$ can be used to represent any of these possible values with an appropriately chosen value of u.

Thanks you for your help solving this, Karan Singh and Archis Welankar! :)

Why does the trigonometric substitution x = a*sec(u) hold for any x, given a is a constant?

1 Answers1