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


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


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


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


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!


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


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


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