# Fiding $\int_0^{2\pi}\frac{\mathrm{d}x}{4\cos^2x+\sin^2x}$

I was checking some old complex analysis homework and I found the following definite integral $$\int_0^{2\pi}\frac{\mathrm{d}x}{4\cos^2x+\sin^2x},$$ had to be found with the residue theorem. Back at the time I thought it was trivial, however I'm trying to do it, but I have no idea on how to star. Could anyone please give me a hint on how to start?

Thanks.

Hint: Let $z=e^{i x}$; $dx=-i dz/z$; $\cos{x}=(z+z^{-1})/2$. The integral is then equal to

$$-i \oint_{|z|=1} \frac{dz}{z} \frac{1}{1+\frac{3}{4} (z+z^{-1})^2}$$

Multiply out, determine the poles, figure out which poles, if any, lie within the unit circle, find the residues of those poles, multiply the sum of those residues (there may only be one, or none) by $i 2 \pi$, and you are done.

• This is what I was looking for, thanks sir. Nov 5 '13 at 17:16

HINT:

Without using Complex Calculus, divide the numerator & the denominator by $\cos^2x$ and substitute $\tan x$ with $u$

• Sorry, I just got your point!!, thanks! Nov 5 '13 at 16:36
• Don't forget the discontinuity in your anti derivative due to the $tanx$. You need to split that integral and use symmetry Nov 5 '13 at 16:42
• @imranfat, agreed. Btw, I have addressed the indefinite part. Nov 5 '13 at 16:44
• @labbhattacharjee I can't use substitution... Nov 5 '13 at 16:53
• @CamiloArosemena, wish I could know the reason behind that. Btw, can you use $$\int\frac{dx}{x^2+a^2}=\frac1a\cdot\arctan \frac xa$$? Nov 5 '13 at 16:57

You can also replace the quadratic trig terms with the double angle formula involving $cos2x$ Your denominator then will consists of a constant and a $cos2x$ term. Then sub away $2x=v$ and then use the Weierstrass substitution converting the integral into a rational function. This is hilarious of course, but what the heck?

if you fcator 4cos^2x from 4cos^2x+sin^2x you will have an arctan form integral at the end note that period of the function is pi so your problem reduce to 4*integral(0 -->pi/2) f(x)

• I'm asked to find the definite integral with the residue theorem... Nov 5 '13 at 16:57
• @khosrotash For some basic information about writing math at this site see e.g. here, here, here and here.
– user93957
Dec 20 '13 at 22:38