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How to evaluate the integral: $$\int_{0}^{\pi}\arctan\left(\frac{1+\cos x}{2+\sin x}\right)dx$$ I didn't get it by partial integration,Any help is appreciated.

I still can't solve this integral. That's a good question. Why doesn't anyone answer it?

JamesJ
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1 Answers1

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Through $\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)$ and symmetry we get that the given integral equals $$ \int_{0}^{\pi/2}\arctan\left(\frac{2+\sin(x)}{2+2\sin(x)}\right)\,dx =\int_{0}^{1}\arctan\left(\frac{2+z}{2+2z}\right)\frac{dz}{\sqrt{1-z^2}}$$ hence it is enough to compute $$ \int_{3/4}^{1}\frac{\arctan(u)\,du}{(2u-1)\sqrt{4u-3}}\quad\text{or}\quad \int_{0}^{1}\arctan\left(\frac{3+v^2}{4}\right)\frac{dv}{1+v^2}.$$ This can be done through integration by parts and the dilogarithms machinery, but the final outcome is not really nice. At least, not the one I got, but maybe I missed some crucial simplification. Anyway, the approach just outlined tends to work pretty well for similar Ahmed-like integrals.

Jack D'Aurizio
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  • I don't really understand this step? – JamesJ Oct 20 '17 at 05:47
  • @JamesJ: which step is troubling you? – Jack D'Aurizio Oct 20 '17 at 09:32
  • What's the connection between the first definite integral and my integral? – JamesJ Oct 20 '17 at 09:58
  • @JamesJ: they are the same. You integral equals $$\int_{0}^{\pi/2}\arctan\frac{1+\cos x}{2+\sin x},dx+\int_{0}^{\pi/2}\arctan\frac{1-\cos x}{2+\sin x},dx = \int_{0}^{\pi/2}\arctan\frac{2+\sin x}{2+2\sin x},dx$$ – Jack D'Aurizio Oct 20 '17 at 10:20
  • But that's what mine is:$\frac{1-\sin x}{2+\cos x}$? – JamesJ Oct 20 '17 at 10:35
  • @JamesJ: this is what you wrote: $$\int_{0}^{\pi}\arctan\left(\frac{1+\cos x}{2+\sin x}\right)dx$$ and there isn't a $\frac{1-\sin x}{2+\cos x}$ in there. Anyway, the above approach applies to that case, too. – Jack D'Aurizio Oct 20 '17 at 10:55
  • My meaning is $\int_{0}^{\pi}\arctan \left(\frac{1+\cos x}{2+\sin x}\right)dx=\int_{0}^{\frac{\pi}{2}}\arctan \left(\frac{1+\cos x}{2+\sin x}\right)dx+\int_{0}^{\frac{\pi}{2}}\arctan \left(\frac{1-\sin x}{2+\cos x}\right)dx$ – JamesJ Oct 20 '17 at 11:05
  • @JamesJ: $\sin(\pi -x)=\sin(x)$ and $\cos(\pi -x)=-\cos(x)$. This is what I mean by symmetry above. – Jack D'Aurizio Oct 20 '17 at 11:07
  • I maybe know what you mean now, thank you very much. – JamesJ Oct 20 '17 at 11:13
  • But how about the last integral? – JamesJ Oct 20 '17 at 11:15
  • @JamesJ: have a look at my answer to this similar question (https://math.stackexchange.com/questions/1359461/how-do-i-evaluate-this-integral-int-frac1-sqrt1-x2-arctan-frac?rq=1) for grasping how to apply the dilogarithms machinery to the computation of the last integral. – Jack D'Aurizio Oct 20 '17 at 11:35