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Find the following integral: $$I=\int_{\pi}^{0}\dfrac{2\cos{t}-8\sin{t}-12\sin{t}\cos{t}}{12\cos^2{t}+16\cos{t}+8}dt$$

My try let $$t=-x$$ $$I=\int_{-\pi}^{0}\dfrac{-2\cos{x}-8\sin{x}-12\sin{x}\cos{x}}{12\cos^2{x}+16\cos{x}+8}dx$$ so $$2I=\left(\int_{\pi}^{0}+\int_{-\pi}^{0}\right)\dfrac{-8\sin{x}-12\sin{x}\cos{x}}{12\cos^2{x}+16\cos{x}+8}dx+\int_{\pi}^{\pi}\dfrac{-2\cos{x}}{12\cos^2{x}+16\cos{x}+8}dx$$

then I can't, thank you.

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  • $\begingroup$ Have you tried the Weierstrass substitution? $\endgroup$ – Potato Oct 18 '13 at 4:30
  • $\begingroup$ Hello,@Potato, you mean $u=\tan{\dfrac{t}{2}}$? But I think is very ugly,I hope see your solution,Thank you $\endgroup$ – china math Oct 18 '13 at 4:33
  • $\begingroup$ Check this technique. $\endgroup$ – Mhenni Benghorbal Oct 18 '13 at 5:15
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    $\begingroup$ As Potato suggested, consider the Weierstrass substitution. It is not so ugly. Do not forget partial fractions. I did it and the result is simply "Pi / 6 + Log[3]" $\endgroup$ – Claude Leibovici Oct 18 '13 at 5:35
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Lucian got it right to start off. The way I see this is to split the integral up:

$$\begin{align}-I &= \int_0^{\pi} dt \frac{\cos{t}}{6 \cos^2{t}+8 \cos{t}+4} + \frac12 \int_0^{\pi} dt \frac{-24 \sin{t} \cos{t} - 16 \sin{t}}{12 \cos^2{t}+16 \cos{t}+8} \\ &= \int_0^{\pi} dt \frac{\cos{t}}{6 \cos^2{t}+8 \cos{t}+4} + \frac12 \left [\log{(12 \cos^2{t}+16 \cos{t}+8)} \right ]_0^{\pi} \\ &= J - \log{3}\end{align}$$

where (borrowing again from Lucian)

$$J = \int_0^{\pi} dt \frac{\cos{t}}{6 \cos^2{t}+8 \cos{t}+4} $$

Now, as is well known in the art of evaluating integrals of this type, we may use the residue theorem. We first exploit the symmetry about $[0, \pi]$ to the full circle and substitute $z=e^{i t}$ such that $dt = -i dz/z$ and get

$$\begin{align}J &= \frac14 \int_0^{2 \pi} \frac{\cos{t}}{3 \cos^2{t}+4 \cos{t}+2} \\ &= -\frac{i}{4} \oint_{|z|=1} \frac{dz}{z} \frac{\frac12 \left (z+z^{-1} \right )}{\frac{3}{4}\left (z^2+z^{-2}+2 \right)+ 2 \left ( z + z^{-1}\right ) + 2} \\&= -\frac{i}{2} \oint_{|z|=1} dz \frac{z^2+1}{3 z^4+8 z^3+14 z^2+8 z+3}\end{align} $$

To evaluate the latter integral, we need to find the poles of the integrand, i.e., the zeroes of the denominator. It may be a quartic, but fortunately it factors:

$$3 z^4+8 z^3+14 z^2+8 z+3 = (3 z^2+2 z+1)(z^2+2 z+3)$$

(I got this by a little trial and error, but it should be clear that such a factoring exists.) From this, we may easily determine the poles:

$$z_1^{\pm} = \frac{-1 \pm i \sqrt{2}}{3}$$ $$z_2^{\pm} = -1 \pm i \sqrt{2}$$

It should also be clear that the poles $z_1^{\pm}$ are inside the unit circle and $z_2^{\pm}$ are outside the unit circle. So now we evaluate the residues only at $z_1^{\pm}$. To compute the residues, use the fact that, for a function $f(z)=p(z)/q(z)$ having a simple pole at $z=z_0$, the residue of $f$ at $z=z_0$ is $p(z_0)/q'(z_0)$. Thus the sum of the residues is

$$\frac{z_1^{+2}+1}{12 z_1^{+3}+24 z_1^{+2} + 28 z_1^++8} + \frac{z_1^{-2}+1}{12 z_1^{-3}+24 z_1^{-2} + 28 z_1^-+8} $$

(Apologies for the confusing notation. The exponents above are not negative, but the minus is part of the variable name.)

What we have now is a mess, but we can clean it up a bit because we know that the poles above are complex conjugates of each other, so that the above sum is really

$$2 \Re{\left [\frac{z_1^{+2}+1}{12 z_1^{+3}+24 z_1^{+2} + 28 z_1^++8} \right ]}$$

Now, I am not going to post an answer here that I could not get by hand. I assure you, it's not as bad as it looks because:

$$z_1^{+} = \frac13 (-1+i \sqrt{2})$$ $$z_1^{+2} = -\frac19 (1+i 2 \sqrt{2})$$ $$z_1^{+3} = \frac{1}{27} (5 + i \sqrt{2})$$

Put this all together to get that (after cancelling twos)

$$\begin{align} \frac{z_1^{+2}+1}{6 z_1^{+3}+12 z_1^{+2} + 14 z_1^++4} &= \frac{\frac19 (8-i 2 \sqrt{2})}{\frac{6}{27} (5 + i \sqrt{2}) - \frac{12}{9} (1+i 2 \sqrt{2}) + \frac{14}{3} (-1 + i \sqrt{2}) + 4} \\ &= \frac{3 (8 - i 2 \sqrt{2})}{6 (5 + i \sqrt{2})-36 (1+i 2 \sqrt{2}) + 126 (-1 + i \sqrt{2}) + 108} \\ &= - \frac{24-i 6 \sqrt{2}}{24-i 60 \sqrt{2}}\\ &= -\frac{324+i 324}{1944}\\&= -\frac{1+i}{6} \end{align}$$

So the sum of the residues is $(-1/6) (-i/2) = i/12$. By the residue theorem, the integral sought, $J$, is $i 2 \pi$ times this sum, or $J=-\pi/6$. Therefore, the original integral is

$$I = \frac{\pi}{6}+\log{3}$$

ADDENDUM

I should add a bit about why we split the integral up in the first place, in case it wasn't clear. I saw pieces that had an antiderivative, and another that did not. It is always best to go after those pieces that may be integrated as usual, using the fundamental theorem, when there is an antiderivative. The other piece that did not, the cosine, could always be done by the residue theorem, even if it is messy. In this case, however, even though there was some work involved, the computation was not at all terrible.

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  • $\begingroup$ Excellento, amigo! $\endgroup$ – Ahaan S. Rungta Oct 18 '13 at 18:15
  • $\begingroup$ @AhaanRungta: ¡Muchas gracias! $\endgroup$ – Ron Gordon Oct 18 '13 at 19:07
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The first thing to notice is that the last two terms of the numerator are nothing else than half of the derivative of the denominator: $$I = \int_\pi^0 \frac{2\cos t + \tfrac12(12\cos^2{t}+16\cos{t}+8)'}{12\cos^2{t}+16\cos{t}+8}dt = \tfrac12\ln(12\cos^2{t}+16\cos{t}+8)\Bigg|_{\ \pi}^{\ 0} + $$ $$ + \int_\pi^0 \frac{\cos t}{6\cos^2{t}+8\cos{t}+4}dt = \ln3 + J$$ where the expression of J is considerably simpler than that of our initial integral I, and can easily be solved by various substitutions, such as the one pointed in the comments above, for instance.

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