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I don´t know how to derive the reduction formula for this integral: $$\int \frac{1}{x^m\sqrt{1-x^2}} \, dx$$

I know you have to use integration by parts, but I have tried everything I can come up with and I don´t get anywhere.

These are some of the ones I've tried:

$u=\frac{1}{x^m},\mspace{0.4cm}\frac{1}{\sqrt{1-x^2}},\mspace{0.4cm} \frac{1}{x^{m+1}},\mspace{0.4cm} \frac{1}{x^{m}\sqrt{1-x^2}}$

$dv=\frac{1}{\sqrt{1-x^2}}, \mspace{0.4cm} \frac{1}{x^m}, \mspace{0.4cm}\frac{x}{\sqrt{1-x^2}}, \mspace{0.4cm} 1 $

I just can't find the trick, somebody please help :(

Hoover Gonzalez
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2 Answers2

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Denote $I_m$ as the integral in the problem. Substitute $\sqrt{1-x^2} = u$, then $\frac{dx}{du} = -\frac{\sqrt{1-x^2}}{x}$ and the integral can be simplified as $$ I_m = -\int\frac{1}{(1-u^2)^{\frac{m+1}{2}}}du \overset{u = \sin{t}}{=} -\int(\sec{t})^mdt.$$ This is simply the reuction formula of $\sec$, which can be found here.

Sam
  • 3,763
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$\textbf{Hint:}$ Add and subtract $x^2$ in the numerator

$$I_m \equiv \int\frac{1-x^2+x^2}{x^m\sqrt{1-x^2}}dx = \int \frac{\sqrt{1-x^2}}{x^m}dx + \int\frac{1}{x^{m-2}\sqrt{1-x^2}}dx$$

Then use integration by parts only on the first term.

Ninad Munshi
  • 37,891