5
$\begingroup$

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

Express

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

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?

=================

EDIT

Continuing @SimplyBeautifulArt's answer:

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

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

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?

$\endgroup$
  • 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. $\endgroup$ – астон вілла олоф мэллбэрг Feb 7 '17 at 1:24
  • $\begingroup$ Agreed with post above $\endgroup$ – mick Feb 7 '17 at 1:25
  • $\begingroup$ @mick Thank you for the confirmation. I was hoping it was not an oddball of a problem. $\endgroup$ – астон вілла олоф мэллбэрг Feb 7 '17 at 1:36
  • $\begingroup$ Holy Jesus good effort $\endgroup$ – Simply Beautiful Art Feb 7 '17 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. $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 12:56
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your following me dude :) $\endgroup$ – mick Feb 7 '17 at 1:40
  • $\begingroup$ Your answer is correct and works but does not actually show the precise answer ... Im curious ! $\endgroup$ – mick Feb 7 '17 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. $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 1:55
  • $\begingroup$ @SimplyBeautifulArt I modified my question by continuing what you started. All good? $\endgroup$ – Ivan Prodanov Feb 7 '17 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.