# Difficult integral.

• Jun 2nd 2010, 05:02 AM
James4321
Difficult integral.
Can any one suggest a solution or method for solving this integration with respect to x

$
\int\frac{1}{\left[a-b\cos(k.x)\right]^{4}}dx
$

Regards
James
• Jun 2nd 2010, 05:28 AM
CaptainBlack
Quote:

Originally Posted by James4321
Can any one suggest a solution or method for solving this integration with respect to x

$
\int\frac{1}{\left[a-b\cos(k.x)\right]^{4}}dx
$

Regards
James

Not elegant, but WolframAlpha gives:

integrate 1&#x2f;&#x28;a-b cos&#x28;k x&#x29;&#x29;&#x5e;4 - Wolfram|Alpha

CB
• Jun 2nd 2010, 09:21 AM
James4321
Quote:

Originally Posted by CaptainBlack

James
• Jun 2nd 2010, 10:18 PM
simplependulum
Quote:

Originally Posted by James4321
Can any one suggest a solution or method for solving this integration with respect to x

$
\int\frac{1}{\left[a-b\cos(k.x)\right]^{4}}dx
$

Regards
James

I'd like to make this more general , let the power to be $n$

To handle such 'troublesome' integral , first thing to do is to reduce it to a simplier form :

$\frac{1}{(-b)^n} \int \frac{dx}{ [ (-a/b) + \cos(kx) ]^n }$

Let $kx = t$ the integral becomes

$\frac{1}{(-b)^n k }\int \frac{dt}{ [A+ \cos(t) ]^n } ~~~ A = -a/b$

it looks better now and we wish to obtain a reduction formula .

Let

$I_n = \int \frac{dt}{ [A+ \cos(t) ]^n }$

$A I_n = \int \frac{(A + \cos(t))-\cos(t)}{ [A+ \cos(t) ]^n } ~dt$

$= I_{n-1} - \int \frac{\cos(t)}{ [A+ \cos(t) ]^n }~dt$

Use integration by parts ,

$= I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n \int \frac{\sin^2(t)}{ [A+ \cos(t) ]^{n+1} } ~dt$

$= I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n \int \frac{1-\cos^2(t)}{ [A+ \cos(t) ]^{n+1} } ~dt$

$= I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - n\int \frac{\cos^2(t)}{ [A+ \cos(t) ]^{n+1} } ~dt$

$= I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - n\int \frac{ [(A+\cos(t)) -A ]^2}{[A+ \cos(t) ]^{n+1} } ~dt$

$=I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - n\int \frac{ (A+\cos(t))^2 -2A(A+\cos(t)) + A^2}{[A+ \cos(t) ]^{n+1} } ~dt$

$= I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - nI_{n-1} +2An I_n - A^2 n I_{n+1}$

Thus

$AI_n = (1 - n) I_{n-1} + 2AnI_n + n(1-A^2) I_{n+1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n }$

$n(1-A^2) I_{n+1} = A(1-2n)I_n + (n-1)I_{n-1} + \frac{\sin(t)}{ [A+ \cos(t) ]^n }$

If $A^2 = 1$ , the reduction only consists of two variables but in general , it consists of three variables : $I_{n-1} , I_n , I_{n+1}$ , that means we have to find $I_1 , I_2$ . By substituting $n=1$ we obtain :

$(1-A^2)I_2 = -A I_1 + 0 + \frac{\sin(t)}{ A+ \cos(t)}$

$I_2 = \frac{A}{A^2-1} I_1 - \frac{\sin(t)}{ ( A^2 -1) [ A + \cos(t) ] }$

I give you the case for $A^2 = 1$, in the above expression , we asumme $A^2 \neq 1$ .

$I_1 = \int \frac{dt}{ A + \cos(t) }$

$= \int \frac{dt}{ A( \cos^2(t/2) + \sin^2(t/2) ) + \cos^2(t/2) - \sin^2(t/2) }$

$= \int \frac{dt}{ (A+1) \cos^2(t/2) +(A-1) \sin^2(t/2) }$

$= \frac{1}{A-1} \int \frac{ \sec^2(t/2) ~dt}{ \tan^2(t/2) + \frac{A+1}{A-1} }$

Sub. $\tan(t/2) = u$

$\sec^2(t/2)~dt = 2du$

$I_1 = \frac{2}{A-1} \int \frac{du}{ u^2 +\frac{A+1}{A-1} }$

If you know the sign of $A^2 - 1 ~~ ( a^2 -b^2 )$ , you should be able to continue ...
• Jun 3rd 2010, 01:05 AM
James4321
Thank you very much for this brilliant suggestion!
• Jun 3rd 2010, 03:20 AM
HallsofIvy
Quote:

Originally Posted by James4321