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.
Exercise:
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
