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I need help in evaluating the following integral, please: $$\int\frac{\cos(x)}{a+b\cos(x)}dx$$ so that we get the following result: $$ \int \frac{\cos x }{a + b\cos x}\:dx = \frac{a}{b\sqrt{a^2-b^2}} \arcsin\left(\frac{b+a\cos x}{a+b\cos x}\right) - \frac1b \arcsin(\cos x) + C $$

Also, I would say I don't have a clue how the substitution can be done. I know another well-known answer which is done using the tangent half-angle substitution, but the result I am asking about exists nowhere online as far as I am concerned. This interesting result is given as a hint to a problem in a physics book and it is completely correct as you can test it yourself. I would appreciate any help. Thanks.

(edit: I came to know that the substitution [u=cos(x)] would lead to the desired formula. So, maybe you can help with that as a hint.)

user170231
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physicist bychoice
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3 Answers3

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Well this probably isn't the way the book does it, but you said you'd appreciate any help. This is how I did it.

$$I=\int\frac{\cos x}{a+b\cos x}dx$$ $$Ib=\int\frac{b\cos x}{a+b\cos x}dx$$ $$Ib=\int\frac{a+b\cos x}{a+b\cos x}dx-a\int\frac{dx}{a+b\cos x}$$ $$Ib=x-a\int\frac{dx}{a+b\cos x}$$ Then we focus on $$J=\int\frac{dx}{a+b\cos x}$$ We may write the integral as $$J=-\int\frac{\sec^2(\frac x2)}{(b-a)\tan^2(\frac x2)-b-a}dx$$ $$J=\frac1{a+b}\int\frac{\sec^2(\frac x2)dx}{\frac{a-b}{a+b}\tan^2(\frac x2)+1}$$ Then we let $$\tan(x/2)=\sqrt{\frac{a+b}{a-b}}u\ \ \Rightarrow\ \ \sec^2(x/2)dx=2\sqrt{\frac{a+b}{a-b}}du$$ Which gives $$J=\frac1{a+b}\int\frac{2\sqrt{\frac{a+b}{a-b}}du}{\frac{a-b}{a+b}\big(\sqrt{\frac{a+b}{a-b}}u\big)^2+1}$$ $$J=\frac2{\sqrt{a^2-b^2}}\int\frac{du}{u^2+1}$$ $$J=\frac2{\sqrt{a^2-b^2}}\arctan u$$ $$J=\frac2{\sqrt{a^2-b^2}}\arctan\bigg[\sqrt{\frac{a-b}{a+b}}\tan\bigg(\frac x2\bigg)\bigg]$$ Hence we have $$Ib=x-\frac{2a}{\sqrt{a^2-b^2}}\arctan\bigg[\sqrt{\frac{a-b}{a+b}}\tan\bigg(\frac x2\bigg)\bigg]$$ Which means $$I=\frac{x}b-\frac{2a}{b\sqrt{a^2-b^2}}\arctan\bigg[\sqrt{\frac{a-b}{a+b}}\tan\bigg(\frac x2\bigg)\bigg]+C$$

clathratus
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    You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :') – physicist bychoice Dec 13 '18 at 15:49
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Hint: Use $$u=\tan \left(\frac{x}{2}\right)$$ so $$\mathrm du=\frac{1}{2}\sec^2 \left(\frac{x}{2}\right)\mathrm dx$$ and $$\sin(x)=\frac{2u}{u^2-1},\cos(x)=\frac{1-u^2}{1+u^2}$$ and $$\mathrm dx=\frac{2 \, \mathrm du}{1+u^2}$$ and we get $$\int\frac{2(1-u^2)}{(u^2+1)^2\left(a+\frac{b(1-u^2)}{u^2+1}\right)}\mathrm du.$$

M. A.
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Dr. Sonnhard Graubner
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  • This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks. – physicist bychoice Dec 12 '18 at 19:12
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The expected solution can be obtained by applying the identity

$$\arctan\sqrt{\dfrac{1-t}{1+t}}=\dfrac12\arccos t=\frac\pi4-\frac12\arcsin t$$

to @clathratus' result, replacing $t=\dfrac{b+a\cos x}{a+b\cos x}=\dfrac{c^2-\tan^2\frac x2}{c^2+\tan^2\frac x2}$ where $c=\sqrt{\dfrac{a+b}{a-b}}$. There is also a careful application of $\sin$/$\arcsin$ on the linear term that leverages the arcsine sum identity.

$$\begin{align*} & \frac xb - \frac{2a}{b\sqrt{a^2-b^2}} \arctan\left(\frac1c\tan\frac x2\right) \\ &= \frac{\color{#c8c8c8}{\frac\pi2} - \arcsin(\cos x)}b + \frac a{b\sqrt{a^2-b^2}} \arcsin \frac{b+a\cos x}{a+b\cos x} \color{#c8c8c8}{- \frac{a\pi}{2b\sqrt{a^2-b^2}}} \end{align*}$$

user170231
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