# How find $\int_{\pi}^{0}\frac{2\cos{t}-8\sin{t}-12\sin{t}\cos{t}}{12\cos^2{t}+16\cos{t}+8}dt$

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.

• Have you tried the Weierstrass substitution? – Potato Oct 18 '13 at 4:30
• Hello,@Potato, you mean $u=\tan{\dfrac{t}{2}}$? But I think is very ugly,I hope see your solution,Thank you – china math Oct 18 '13 at 4:33
• Check this technique. – Mhenni Benghorbal Oct 18 '13 at 5:15
• 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]" – Claude Leibovici Oct 18 '13 at 5:35

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

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

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.