I'm self learning calculus and I stumbled upon the following problem:


$I_n =\int \frac{dx}{(x^2+a^2)^n}$

Using $I_{n-1}$ ($a$ is a positive parameter and $n=2,3,4,...$)

Is this about double integrals? Could anyone please elaborate a bit more so I can learn how to solve this type of problems?



Continuing @SimplyBeautifulArt's answer:

$I_n=a^{1-2n}\int{cos^{2n-2}(u)du} = a^{1-2n}\int{cos^{2n-3}(u)cos(u)du}$


Integrating by parts($f:cos^{2n-3}(u); dg:cos(u)du$):

$I_n=a^{1-2n}(cos^{2n-3}(u)sin(u) + (2n-3)\int{cos^{2n-4}(u)sin^2(u)du})$

$I_n=a^{1-2n}(cos^{2n-3}(u)sin(u) + (2n-3)(\int{cos^{2n-4}(u)du} -\int{cos^{2n-2}(u)du}))$

$I_n=a^{1-2n}cos^{2n-3}(u)sin(u) + (2n-3)(\frac{a^{3-2n}}{a^2}\int{cos^{2n-4}(u)du} -a^{1-2n}\int{cos^{2n-2}(u)du})$

$I_n=a^{1-2n}cos^{2n-3}(u)sin(u) + (2n-3)(\frac{I_{n-1}}{a^2} -I_n)$

$I_n=(a^{1-2n}cos^{2n-3}(u)sin(u) + (2n-3)\frac{I_{n-1}}{a^2})/2n-2$

Recall $u=arctan(\frac xa)$

$I_n=(a^{1-2n}\frac{1}{\sqrt{1+(x/a)^2}}^{2n-3}\frac{x/a}{\sqrt{1+(x/a)^2}} + (2n-3)\frac{I_{n-1}}{a^2})/2n-2$

Is that all?

  • 2
    $\begingroup$ Recursion type problem in integration. Typically involving use of integration by parts. In this case, I would first try a trigonometric substitution before using integration by parts. $\endgroup$ Feb 7, 2017 at 1:24
  • $\begingroup$ Agreed with post above $\endgroup$
    – mick
    Feb 7, 2017 at 1:25
  • $\begingroup$ @mick Thank you for the confirmation. I was hoping it was not an oddball of a problem. $\endgroup$ Feb 7, 2017 at 1:36
  • $\begingroup$ Holy Jesus good effort $\endgroup$ Feb 7, 2017 at 12:52
  • $\begingroup$ So. I can't actually fully see everything here (mobile) so I'll look it over when I can. Note that you can also ask WolframAlpha for the answer. $\endgroup$ Feb 7, 2017 at 12:56

1 Answer 1


Use the substitution $x=a\tan(u)$ to get

$$I_n=\int\frac{a\sec^2(u)}{(a^2\tan^2(u)+a^2)^n}\ du$$

Recall the trigonometric identity $1+\tan^2=\sec^2$ to reduce this to

$$I_n=a^{1-2n}\int\sec^{2-2n}(u)\ du$$

$$=a^{1-2n}\int\cos^{2n-2}(u)\ du$$

This is then handled using pythagorean theorem, integration by parts, and/or substitution, depending on the value of $n$, as described in this post.

  • $\begingroup$ Your following me dude :) $\endgroup$
    – mick
    Feb 7, 2017 at 1:40
  • $\begingroup$ Your answer is correct and works but does not actually show the precise answer ... Im curious ! $\endgroup$
    – mick
    Feb 7, 2017 at 1:42
  • $\begingroup$ @mick Oh, sorry, my bad. :P I will say, "closed form" may be found via Euler's formula and binomial expansion, though I think this is worse than providing a method. $\endgroup$ Feb 7, 2017 at 1:55
  • $\begingroup$ @SimplyBeautifulArt I modified my question by continuing what you started. All good? $\endgroup$ Feb 7, 2017 at 12:14

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