More on the three-fold way

A while ago I posted about what seems to be a simple calculus problem---to find the area of a circle by integration. In fact we looked at three ways, and two were easy. But the first way is harder, and that involves y as a function of x with the goal to integrate y(x) from x = -R to x = R. At the time, I solved the problem by numerical integration.

The function I could not integrate is

y = √(R2 - x2)

or sticking with a unit circle:

y = √(1 - x2)

Reading further in Strang's Calculus, I find that there is a way to do it. The method has two parts: (i) a trigonometric substitution and (ii) integration by parts, which is a reversal of the product rule for differentiation.

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Integration by parts

If we have two functions of x, u and v, and we want d/dx of uv, by the product rule we get:

d/dx uv = u dv/dx + v du/dx

Thinking about integrating this (without the x's):

uv = ∫u dv + ∫v du ∫u dv = uv - ∫v du

So the trick is, when given a difficult integral, to try to imagine it transformed into ∫u dv. If ∫v du is easy or just easier, we have helped ourselves. Suppose we have:

∫cos2 x dx ∫cos x cos x dx

Then let

u = cos x dv = cos x dx v = sin x du = -sin x dx ∫cos2 x dx = = ∫u dv = uv - ∫v du = sin x cos x + ∫sin x sin x dx

And a trick:

∫cos2 x dx = = sin x cos x + ∫(1-cos2 x) dx = sin x cos x + x - ∫cos2 x dx

And another trick!:

2 ∫cos2 x dx = sin x cos x + x ∫cos2 x dx = 1/2(sin x cos x + x)

Amazing. And:

∫sin2 x dx = = ∫ (1 - cos2 x) dx = x - 1/2(sin x cos x + x) = 1/2(x - sin x cos x)

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The "double angle" method

I quoted Strang the other day that:

sin(s + t) = sin s cos t + cos s sin t cos(s + t) = cos s cos t - sin s sin t

These formulas lead to a pretty straightforward derivation of the integral of
cos² t (happily, we get the same result as above):

if s = t then we have:

sin 2t = 2 sin t cos t cos 2t = cos2 t - sin2 t = cos2 t - 1 + cos2 t = 2 cos2 t - 1 cos2 t = 1/2 (1 + cos 2t)

Substitute:

1 - sin2 t = 1/2(1 + cos 2t) sin2 t = 1 - 1/2(1 + cos 2t) = 1/2(1 - cos 2t)

We can integrate that easily:

∫sin2 t dt = = ∫[ 1/2 - 1/2(cos 2t)] dt = t/2 - 1/4(sin 2t) = t/2 - 1/2(sin t cos t) = 1/2(t - sin t cos t)

and

∫cos2 t dt = = 1/2 ∫(1 + cos 2t) dt = 1/2 (t + 1/2 sin 2t) = 1/2 (t + sin t cos t)

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But wait, we haven't solved the problem. We want to integrate:

dy = √(1 - x2)

substitute

x = sin θ dx = cos θ dθ

then we have

y = ∫ √(1 - sin2 θ) cos θ dθ y = ∫ cos θ cos θ dθ y = ∫ cos2 θ dθ = 1/2(θ + sin θ cos θ)

change back to x

y = 1/2 [ sin-1 x + x √(1-x2) ]

(Notice it's the inverse sine). Now, evaluate between x = 0 and x = 1:

y = 1/2[ π/2 + 0 - 0 - 0] = π / 4

I still haven't worked out the R² part.