# A Simple Definite Integral

1. Apr 2, 2009

### bjgawp

I have a problem which involves finding the moment of inertia which involves evaluating: $$\int_0^1 x^2 \left(\frac{\pi}{2} - \sin^{-1} x\right)dx$$

Now eventually, we end up having to evaluate the integral through integration by parts: $$\int_0^1 \frac{x^3}{\sqrt{1-x^2}} \ dx$$

Now this is really an improper integral with the problem at $$x= 1$$ so we really have: $$\lim_{t \to 1^-} \int_0^t \frac{x^3}{\sqrt{1-x^2}} dx$$

Now the problem that has occurred to me is that I'm getting two different results through two different methods.

Method 1: A u-substitution

Let: $$u = \sqrt{1-x^2} \ \Rightarrow \ - u du = x dx$$

$$x = 0 \ \Rightarrow \ u = 1$$, and $$x \to 1 \ \Rightarrow \ u \to 0$$

So we have the new integral: $$\lim_{s \to 0^+} \int_s^1 \frac{1-u^2}{u} \ du = \lim_{s \to 0^+} \left(\ln u - \frac{u^2}{2}\right) \Bigg|_s^1$$

which diverges.

Method 2: A trig-sub

Let: $$x = \sin \theta \ \Rightarrow \ dx = \cos \theta d \theta$$

So, the integral becomes: $$\lim_{s \to \frac{\pi}{2}} \int_0^s \sin^3 \theta d \theta$$

but this is no longer improper and we do get a result.

So what's going on here? If we do the trig-sub, we get the correct answer to the original inertia question but if we look at the function $$f(x) = \frac{x^3}{\sqrt{1-x^2}}$$, it should diverge!

2. Apr 2, 2009

### Hurkyl

Staff Emeritus
It's hard to know without seeing your arithmetic, but I think you dropped the factor in red.

3. Apr 2, 2009

### bjgawp

Oh whoops! Wow what a silly mistake. Thanks!