Integral follow-up

One HUGE reason to ‘blog is to get feedback on your ideas.  Thanks to David’s response to my last post on my attempts to integrate \int x^2\sqrt{x^2+1} dx, I got another lead on how to tackle the problem –and– a little (comforting) confirmation that the integral really was difficult.

The suggestion that integrals of odd powers of secant might cycle was all I needed.  I ignoring the hint that Wikipedia might hold a solution to find a way for myself. Here’s what I found.

To compute I=\int sec^3 \theta d\theta, let

Integral2-5

One application of Integration by Parts, a trig identity, and a cycling integral gives

Integral2-1

Similarly, to compute I=\int sec^5 \theta d\theta, let

Integral2-6

Using the result of \int sec^3\theta d\theta, Integration by Parts once, a trig identity, and a cycling integral gives this solution.

Integral2-2

From the first post, if x=tan\theta, then \int x^2\sqrt{x^2+1} dx = \int tan^2\theta sec^3\theta d\theta. A Pythagorean trig identity turns this into a linear combination of the previous two integrals.  The problem that initially stumped me has now been solved using using only circular trig and other integration techniques!

Integral2-3

With potential domain restrictions, the initial substitution x=tan\theta with a Pythagorean trig identity gives sec\theta=\sqrt{x^2+1}.  Therefore,

Integral2-4

one of the original solution forms, found this time completely independent of hyperbolic trig.  Thanks for the suggestion, David!

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