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I have to find the recursive formula for this expression : $\int\dfrac{dx}{(a+b\cos x)^n}$ .

This is all what I've done :

$$\int\frac{dx}{(a+b\cos x)^n} = \frac{1}{a}\int\frac{adx}{(a+b\cos x)^n}=\frac{1}{a}\int\frac{a+b\cos x-b\cos x}{(a+b\cos x)^n}dx=\frac{1}{a}I_{n-1}-\frac{b}{a}\int\frac{\cos xdx}{(a+b\cos x)^n} .$$

Now I have to solve :

$\int\dfrac{\cos xdx}{(a+b\cos x)^n}$

Harry Peter
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zanita
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  • I don't know if this will lead to a solution, but if that cos were sin, then this would be easy by spotting that the integrand came from differentiating some constant multiple of $(a+b \cos x)^{-n+1}$. If we differentiate the integral with respect to $b$, then we get $\cos^2 = 1-\sin^2$ on the numerator, and the $\sin^2$ is nearly what we want… – Patrick Stevens Nov 19 '16 at 14:35

1 Answers1

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$\int\dfrac{dx}{(a+b\cos x)^n}$

$=\dfrac{1}{a}\int\dfrac{a}{(a+b\cos x)^n}~dx$

$=\dfrac{1}{a}\int\dfrac{a+b\cos x-b\cos x}{(a+b\cos x)^n}~dx$

$=\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}-\dfrac{b}{a}\int\dfrac{\cos x}{(a+b\cos x)^n}~dx$

$=\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}-\dfrac{b}{a}\int\dfrac{d(\sin x)}{(a+b\cos x)^n}$

$=-\dfrac{b\sin x}{a(a+b\cos x)^n}+\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}+\dfrac{b}{a}\int\sin x~d\left(\dfrac{1}{(a+b\cos x)^n}\right)$

$=-\dfrac{b\sin x}{a(a+b\cos x)^n}+\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}+\dfrac{b^2n}{a}\int\dfrac{\sin^2x}{(a+b\cos x)^{n+1}}~dx$

$=-\dfrac{b\sin x}{a(a+b\cos x)^n}+\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}-\dfrac{b^2n}{a}\int\dfrac{\cos^2x-1}{(a+b\cos x)^{n+1}}~dx$

$=-\dfrac{b\sin x}{a(a+b\cos x)^n}+\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}-\dfrac{n}{a}\int\dfrac{b^2\cos^2x-b^2}{(a+b\cos x)^{n+1}}~dx$

$=-\dfrac{b\sin x}{a(a+b\cos x)^n}+\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}-\dfrac{n}{a}\int\dfrac{b^2\cos^2x+2ab\cos x+a^2-2ab\cos x-a^2-b^2}{(a+b\cos x)^{n+1}}~dx$

$=-\dfrac{b\sin x}{a(a+b\cos x)^n}+\dfrac{1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}-\dfrac{n}{a}\int\dfrac{(a+b\cos x)^2-2a(a+b\cos x)+a^2-b^2}{(a+b\cos x)^{n+1}}~dx$

$=-\dfrac{b\sin x}{a(a+b\cos x)^n}-\dfrac{n-1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}+2n\int\dfrac{dx}{(a+b\cos x)^n}-\dfrac{(a^2-b^2)n}{a}\int\dfrac{dx}{(a+b\cos x)^{n+1}}$

$\therefore\dfrac{(a^2-b^2)n}{a}\int\dfrac{dx}{(a+b\cos x)^{n+1}}=-\dfrac{b\sin x}{a(a+b\cos x)^n}+(2n-1)\int\dfrac{dx}{(a+b\cos x)^n}-\dfrac{n-1}{a}\int\dfrac{dx}{(a+b\cos x)^{n-1}}$

$\int\dfrac{dx}{(a+b\cos x)^{n+1}}=-\dfrac{b\sin x}{(a^2-b^2)n(a+b\cos x)^n}+\dfrac{a(2n-1)}{(a^2-b^2)n}\int\dfrac{dx}{(a+b\cos x)^n}-\dfrac{n-1}{(a^2-b^2)n}\int\dfrac{dx}{(a+b\cos x)^{n-1}}$

Hence $\int\dfrac{dx}{(a+b\cos x)^n}=-\dfrac{b\sin x}{(a^2-b^2)(n-1)(a+b\cos x)^{n-1}}+\dfrac{a(2n-3)}{(a^2-b^2)(n-1)}\int\dfrac{dx}{(a+b\cos x)^{n-1}}-\dfrac{n-2}{(a^2-b^2)(n-1)}\int\dfrac{dx}{(a+b\cos x)^{n-2}}$

Harry Peter
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