reduction formulae question help!! (1 Viewer)

poptarts12345 · Mar 17, 2021

how do i do this question?

quickoats · Mar 17, 2021

poptarts12345 said:
View attachment 30396

how do i do this question?

Try using IBP on I_(n-1) where u=1/(x^2+1)^(n-1) and dv=1. (Haven’t done the qn but it seems like it may work). Note that doing IBP on I_n will most likely give you an (x^2+1)^(n+1) on the denominator. With this question it looks tempting to do tan inverse but it doesn’t appear in the reduced form.

Letting dv=1 is a good trick to use in general

Qeru · Mar 17, 2021

poptarts12345 said:
View attachment 30396

how do i do this question?

$\frac{d}{dx}\left(\frac{x}{(x^2+1)^{n-1}}+(2n-3)I_{n-1}\right)$

$=\frac{1-(2n-3)x^2}{(x^2+1)^{n}}+\frac{2n-3}{(x^2+1)^{n-1}}$

$=\frac{1-(2n-3)x^2}{(x^2+1)^{n}}+\frac{(2n-3)(x^2+1)}{(x^2+1)^{n}}$

$=\frac{2n-2}{(x^2+1)^{n}}$

$\therefore \int \frac{1}{(x^2+1)^{n}}=\frac{1}{2(n-1)}\int \frac{d}{dx}\left(\frac{x}{(x^2+1)^{n-1}}+(2n-3)I_{n-1}\right)$

$=\frac{1}{2(n-1)}\left(\frac{x}{(x^2+1)^{n-1}}+(2n-3)I_{n-1}\right)$

reduction formulae question help!! (1 Viewer)

poptarts12345

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