Integration by Reduction

Hi wondered if anyone had a list of some common integration by reduction questions and what to split them into eg sec^n x should be split into sec^2 x sec^(n-2) x

Reply 1

davros

Original post by apolaroidofus

Hi wondered if anyone had a list of some common integration by reduction questions and what to split them into eg sec^n x should be split into sec^2 x sec^(n-2) x

There probably won't be a definitive list since this is a very specific topic and only comes up infrequently in exams, but googling "integration reduction formula" should give you plenty of examples of how this technique is used in practice :smile:

Reply 2

DFranklin

When integrating by parts you need to split the integrand into something you can differentiate (easy), and something you can integrate which doesn't get more complicated when you do so (and whose integrand doesn't look totally unlike what you already have), which is harder. In general, there aren't many options for what you're going to integrate once you think about it.

The one reasonably common one people struggle with is

\displaystyle I_n = \int \dfrac{1}{(1+x^2)^n}\,dx

, which needs two non-obvious steps. The first trick is to IBP with "v" =

\dfrac{1}{(1+x^2)^n}

and "du" = 1. The second thing that throws people here is that then after rearranging you end up with a

I_{n+1}

term instead of a

I_{n-1}

term. But you can rearrange/relabel to get a more traditional reduction form. [e.g. if you know

I_{n} = \dfrac{n}{n+1} I_{n+1}

, you can rewrite this as

I_n = \dfrac{n}{n-1} I_{n-1}

]

Couple of other tricks to be aware of:

Integrand of form: x^n f(x^k). Consider whether it's viable to integrate

x^{k-1}f(x^k)

via the substitution u = x^k (or by recognition). E.g. with

\int x^n e^{x^2}\,dx

you want to take "du" =

x e^{x^2}

It is sometimes the case that a reduction formula works, but brute force algebra is quicker. The most common case is

\int \sin^n x \,dx

with n known and odd, where it's quicker to rewrite as

\int \sin x (1-\cos^2x)^{(n-1)/2} \,dx

and integrate by recognition.

E.g.

\int \sin^7 x \,dx = \int \sin x(1-cos^2 x)^3

= \int \sin x - 3 \sin x \cos^2 x + 3 \sin x \cos^4 x - \sin x \cos^6 x\,dx

= -\cos x + \cos^3 x - \frac{3}{5} \cos^5 x + \frac{1}{7} \cos^7 x + C

(edited 1 year ago)

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Integration by Reduction

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