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
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/(a-b cos(k x))^4 - Wolfram|Alpha
CB
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 ...