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I'm trying to evaluate the following integral with a Weierstrass substitution:

$$ \int_{\pi/3}^{4\pi/3} \frac{3}{13 + 6\sin x - 5\cos x} \text{d}x $$

This comes out to about 0.55 when evaluated numerically. I thought it would be possible to rewrite the integral using the substitution $t = \tan\frac{x}{2}$ like so:

$$ \text{Upper limit: }\frac{4\pi}{3} \to \tan\frac{4\pi}{6} \to -\sqrt{3} $$

$$ \text{Lower limit: }\frac{\pi}{3} \to \tan\frac{\pi}{6} \to \sqrt{3}/3 $$

$$ \int_{\sqrt{3}/3}^{-\sqrt{3}} \frac{3(1+t^2)}{2(3t+1)^2 + 6} \cdot \frac{2}{1 + t^2} \text{d}x $$

However, evaluating this numerically now gives -1.26, so it's clear that this method is wrong. I am sure that

$$ \frac{3}{13 + 6\sin x - 5\cos x} = \frac{3(1+t^2)}{2(3t+1)^2 + 6} $$ So something must have gone wrong changing the limits, but I've never come across this problem before. I thought the integral may be improper but plotting $\frac{3}{13 + 6\sin x - 5\cos x}$ on a graph shows it has no asymptotes or any other obvious problems in the range of integration.

Any help would be greatly appreciated :)

Mittens
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Olly Britton
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2 Answers2

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Take care of the discontinuity at $x=\pi$ with the half angle substitution $t=\tan\frac x2$

\begin{align} &\int_{\pi/3}^{4\pi/3} \frac{3}{13 + 6\sin x - 5\cos x} dx \overset{t=\tan\frac x2}=\bigg(\int_{\frac1{\sqrt3}}^\infty +\int_{-\infty}^{-\sqrt3}\bigg) \frac3{4+6t+9t^2}dt \end{align}

Quanto
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The Weierstrass substitution is valid for $x\in(-\pi,\pi)$ or $x\in(\pi,3\pi)$, but not for an interval containing a neighborhood of $\pi$. As $x$ tends to $\pi$ from below, $z$ tends to $+\infty$. As $x$ tends to $\pi$ from above, $z$ tends to $-\infty$. $$ \begin{align} &\int^{4\pi/3}_{\pi/3}\frac3{13+6\sin(x)-5\cos(x)}\,\mathrm{d}x\\ &=\int^{\pi}_{\pi/3}\frac3{13+6\sin(x)-5\cos(x)}\,\mathrm{d}x +\int^{4\pi/3}_{\pi}\frac3{13+6\sin(x)-5\cos(x)}\,\mathrm{d}x\tag{1a}\\ &=\int_{1/\sqrt3}^\infty\frac3{13+6\frac{2z}{1+z^2}-5\frac{1-z^2}{1+z^2}}\frac{2\,\mathrm{d}z}{1+z^2} +\int_{-\infty}^{-\sqrt3}\frac3{13+6\frac{2z}{1+z^2}-5\frac{1-z^2}{1+z^2}}\frac{2\,\mathrm{d}z}{1+z^2}\tag{1b}\\ &=\int_{1/\sqrt3}^\infty\frac{3\,\mathrm{d}z}{4+6z+9z^2} +\int_{-\infty}^{-\sqrt3}\frac{3\,\mathrm{d}z}{4+6z+9z^2}\tag{1c}\\ &=\int_{\tan^{-1}\left(\frac{3+\sqrt3}3\right)}^{\pi/2}\frac{\mathrm{d}u}{\sqrt3} +\int_{-\pi/2}^{\tan^{-1}\left(\frac{-9+\sqrt3}3\right)}\frac{\mathrm{d}u}{\sqrt3}\tag{1d}\\[3pt] &=\frac1{\sqrt3}\tan^{-1}\left(\frac{60-24\sqrt3}{13}\right)\tag{1e} \end{align} $$ Explanation:
$\text{(1a)}$: break the integral into ranges where $z$ is increasing
$\text{(1b)}$: $z=\tan(x/2)$, $\sin(x)=\frac{2z}{1+z^2}$, $\cos(x)=\frac{1-z^2}{1+z^2}$, $\mathrm{d}x=\frac{2\,\mathrm{d}z}{1+z^2}$
$\text{(1c)}$: simplify
$\text{(1d)}$: if $\sqrt3z+1/\sqrt3=\tan(u)$, then $4+6z+9z^2=3\sec^2(u)$ and $\mathrm{d}z=\frac1{\sqrt3}\sec^2(u)\,\mathrm{d}u$
$\text{(1e)}$: $\tan^{-1}\left(\frac{3+\sqrt3}3\right)\gt\frac\pi4$ and $\tan^{-1}\left(\frac{9-\sqrt3}3\right)\gt\frac\pi4$, and
$\phantom{\text{(1e): }}$$\tan\left(\pi-\tan^{-1}\left(\frac{3+\sqrt3}3\right)-\tan^{-1}\left(\frac{9-\sqrt3}3\right)\right)=\frac{60-24\sqrt3}{13}$

robjohn
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