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

or sticking with a unit circle:

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.

### 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:

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

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:

Then let

And a trick:

And another trick!:

Amazing. And:

### The "double angle" method

I quoted Strang the other day that:

These formulas lead to a pretty straightforward derivation of the integral of

cos

^{2}t (happily, we get the same result as above):

if s = t then we have:

Substitute:

We can integrate that easily:

and

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

substitute

then we have

change back to x

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

I still haven't worked out the R

^{2}part.

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