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How do I evaluate $\int_0^{\pi}\frac{1}{5+3\cos{x}}dx$? Because for this type of integral I would normally use the substitution $t=\tan{\frac{x}{2}}$ but $\tan{\frac{\pi}{2}}$ is undefined. How can I evaluate this integral still using this method (I want to still use this method because it is the method my school expects me to use)?

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    $\begingroup$ Break it in two and use limits... $\endgroup$
    – abiessu
    Sep 24, 2021 at 4:38
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    $\begingroup$ Note that $\lim_{x\to\pi^-}t=+\infty$, so you get $\int_0^{+\infty}\dots\,\mathrm{d}t$, etc. $\endgroup$ Sep 24, 2021 at 4:45
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    $\begingroup$ @Quippy I'll have to think about this : I'm also thinking about whether the $+\infty$ causes a problem at all. You see, forgetting about the limits of the integral which are $0$ and $\pi$, the indefinite integral $\int \frac 1{5 + 3 \cos x}$ can be evaluated using the substitution $ u= \tan \frac x2$ to equal $\frac{-1}{2} \tan^{-1}(2 \cot(\frac x2))$. The point is that $0$ and $\pi$ can be substituted into this anti-derivative , even if there's infinities lurking, because $\tan^{-1}(\infty) = \frac \pi 2$. For a 2 mark question I'd say realizing this reflects the marking scheme better. $\endgroup$ Sep 24, 2021 at 5:30
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    $\begingroup$ @TeresaLisbon If it helps, they gave you the graph of $y=\frac{1}{5+3\cos{x}}$, I tried to use symmetry by instead evaluating up to the limit of $2\pi$ then having it but when I substituted those limits it came out as $0$ because I left my indefinite form as $\frac12\tan^{-1}\left(\frac12\tan\left(\frac{x}{2}\right)\right)$. $\endgroup$ Sep 24, 2021 at 5:48
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    $\begingroup$ @Quippy The graph doesn't help get the exact value, beyond the fact that it might show some symmetries (the same symmetries that I used to create my earlier decomposition, if you like). I think Jean-Claude's link is the key that explains why the $2 \pi$ issue arises. Basically, your substitution has to be a bijective substitution on the interval, and $\tan \frac x2$ has this property on $[0,\pi]$ but not $[0,2 \pi]$. So that's something to keep in mind. I think the hacky method, unfortunately, seems to be the 2-mark method here. $\endgroup$ Sep 24, 2021 at 6:05

2 Answers 2

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Use $\cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$, then $$I=\int_{0}^{\pi} \frac{dx}{5+3\cos x}=\int_{0}^{\pi}\frac{\sec^2(x/2) dx}{8+2\tan^2(x/2)}.$$ $\tan(x/2)=t \implies \sec^2(x/2)dx=2dt$ $$I=\int_{0}^{\infty} \frac{dt}{4+ t^2}=\frac{1}{2} \tan^{-1} t/2|_0^{\infty}=\frac{\pi}{4}.$$

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Self answer after comment discussion;

$\int_0^{\pi}\frac{1}{5+3\cos{x}}dx=\int_0^{+\infty}\frac{1}{5+3\left(\frac{1-t^2}{1+t^2}\right)}\frac{2}{1+t^2}dt$ after the substitution of $t=\tan\left(\frac{x}{2}\right)$
$=\int_0^{+\infty}\frac{2}{8+2t^2}dt$

$=\sqrt{2}\int_0^{+\infty}\frac{\sqrt{2}}{(2\sqrt{2})^2+(\sqrt2t)^2}dt$

$=\frac{\sqrt{2}}{2\sqrt2}\tan^{-1}\left(\frac{\sqrt2\times\infty}{2\sqrt2}\right)-\frac{\sqrt{2}}{2\sqrt2}\tan^{-1}\left(\frac{\sqrt2\times0}{2\sqrt2}\right)$

$=\frac12\tan^{-1}{(+\infty)}$

$=\frac{\pi}{4}\because\tan^{-1}{(+\infty)}=\frac{\pi}{2}$

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