Integration 3

I want to integrate (a²-x²)-1/2. I’m odd like that. The term inside the brackets is screaming at me that Pythagoras applies, so I draw a little sketch of a triangle with a hypotenuse of length a and decide I’d like to call sin θ = x/a.

This is going to become a substitution. I am going to switch the integral to work in θ rather than x.

Since I have declared sin θ = x/a, then x = a sin θ, and I should look at (a²-x²), which becomes a²(1-sin²θ) = a²cos2θ. 

Similarly I will need to look at dx/dθ = a cosθ, or, to write it differently, dx = a cosθ dθ ;

I shall be replacing dx with that expression.

So my integral ∫(a2-x2)-1/2 dx becomes

∫ (a²cos2θ) -1/2  . a cosθ dθ = ∫ dθ.

This is so very simple that students often can’t do the integration, which comes to just θ (plus a possible constant of integration). 

Inverting the substitution, I have ∫ (a²-x²)-1/2 = sin-1 (x/a) + c.

Got the idea?

• Pick a substitution and declare it (“Let x = a sinθ”), 
• convert any strange expressions within the function from the old variable to the new and
• clearly show what happens to the old differential (the dx bit). If your integral is definite (it has limits of perhaps a and b) then 
• you might want to also change the limits from x=a and x=b  to θ=θ0 and θ=θ1. Often this is a decision you take later, after you have shown that your substitution is helpful.


1   Do ∫ (a²-x²)-1/2 dx  using the substitution x = a cos θ

2   Do ∫ (a²+x²)-1 dx  using the substitution x = a tan θ

3   Do ∫ e1+cos θ sin x dx  using the substitution u = 1 + cos θ

I find many students (claim to) understand the principle here but fail to write down all that they need to write down in two senses: 
  (i) they don’t write enough to help them reach a correct result; 
  (ii) they also don’t write enough to enable an exam marker to award method marks. It is as if they need a formal layout to follow: that seems to demonstrate a failure to understand the substitution process. Hence this page.

DJS 20130430



© David Scoins 2017