# Thread: Difficult integral.

1. ## Difficult integral.

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

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

Regards
James

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

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

Regards
James
Not elegant, but WolframAlpha gives:

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

CB

3. Originally Posted by CaptainBlack
Thank you Giordano for your prompt reply. I tried this already. Unfortunately, WolframAlpha result is wrong.

James

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

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

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

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

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

Let $\displaystyle kx = t$ the integral becomes

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

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

Let

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

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

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

Use integration by parts ,

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

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

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

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

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

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

Thus

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

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

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

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

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

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

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

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

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

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

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

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

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

If you know the sign of $\displaystyle A^2 - 1 ~~ ( a^2 -b^2 )$ , you should be able to continue ...

5. Thank you very much for this brilliant suggestion!

6. Originally Posted by James4321
Thank you Giordano for your prompt reply. I tried this already. Unfortunately, WolframAlpha result is wrong.

James
"Giordano" is not Captain Black's name. Giordano Bruno is the author of the quote he gives in his signature.

7. Sorry about this! I would also like to thank Simplependulum for his elegant and complete solution.

James4321