What's the difference between Recursion & Reduction in terms of Integration?

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Homework Statement



My book talks about the "reduction formulas" for evaluating trigonometric integrals by parts. However, is this the same thing as "recursive" formulas for integration by parts, a term which is not mentioned in my calculus book?


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The Attempt at a Solution

 
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Recursive means something defined in terms of itself, but with a different conditions. For example, factorial: n! = (n - 1)!n.

It is not entirely clear what in you book is called "reduction formulas", but if that means, for example, something with degrees defined in terms of something similar with smaller degrees, then this could be equally called recursive formulas.
 
voko said:
Recursive means something defined in terms of itself, but with a different conditions. For example, factorial: n! = (n - 1)!n.

It is not entirely clear what in you book is called "reduction formulas", but if that means, for example, something with degrees defined in terms of something similar with smaller degrees, then this could be equally called recursive formulas.

Yes, they are talking about degrees defined in terms of something similar with smaller degrees when mentioning "reduction formulas." Thanks for clarifying and confirming they are different words meaning the same thing in this case.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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