I'm not really unsure of how to approach this problem. I was thinking of reparametrizing the sin and the cos to its exponential form but I realize that it becomes even messier and leads sort of nowhere. There are no singularities for this function f(x) I believe, so there's not really a way to use the Residue theorem either. Can anyone give me some help on this?
Asked
Active
Viewed 3,267 times
1
-
1Did you try dropping $\sin^2x$ into double angle formula for $\cos x$? – abiessu May 09 '15 at 21:11
-
Yea, I think it gets really nasty as well. – cambelot May 09 '15 at 21:14
-
I would go with half-angle tangent substitution, but I'm too tired to do this now. – Wojowu May 09 '15 at 21:19
-
Can you show me how that might work out? I guess it might be a mistake on my end. – cambelot May 09 '15 at 21:26
-
there will actually be singularities on the complex plane $5+3cos(z)=0$ does have a solution. – WW1 May 09 '15 at 21:43
-
possible duplicate of Evaluating $\int P(\sin x, \cos x) \text{d}x$ – epimorphic May 09 '15 at 22:11
-
@epimorphic A definite integral over $[0,2\pi]$ can be computed in ways that do not require an antiderivative. – May 10 '15 at 03:41
4 Answers
2
Hint:
Using half-angle tangent substitution $u=\tan(\dfrac{x}{2})$ and with some algebra you find: $$ 4\int \dfrac{u^2}{u^6+6u^4+9u^2+4} dx =4\int \dfrac{u^2}{(u^2+1)^2(u^2+4)} dx $$
Now use partial fractions.
Emilio Novati
- 62,675
2
Suppose we seek to evaluate $$\int_0^{2\pi} \frac{\sin^2 x}{5+3\cos x} dx.$$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) \; dx$ and hence $\frac{dz}{iz} = dx$ to obtain $$\int_{|z|=1} \frac{(z-1/z)^2/4/(-1)}{5+3/2(z+1/z)} \frac{dz}{iz} \\ = -\int_{|z|=1} \frac{(z-1/z)^2}{20+6(z+1/z)} \frac{dz}{iz} \\ = -\int_{|z|=1} \frac{z^2-2+1/z^2}{20+6z+6/z} \frac{dz}{iz} \\ = -\int_{|z|=1} \frac{z^4-2z^2+1}{20z^2+6z^3+6z} \frac{dz}{iz} \\ = -\frac{1}{i} \int_{|z|=1} \frac{1}{z^2} \frac{z^4-2z^2+1}{6z^2+20z+6} \; dz.$$
There are three poles, one at $z=-3$, another one at $z=-1/3$ and another one at $z=0.$ Only the latter two contribute.
The pole at $z=-1/3$ is simple and hence the residue is $$\left.\frac{1}{z^2} \frac{z^4-2z^2+1}{12z+20}\right|_{z=-1/3} = \frac{4}{9}.$$
The remaining contribution is from $$\int_{|z|=1} \frac{1}{z^2} \frac{1}{6z^2+20z+6} \; dz =\frac{1}{6} \int_{|z|=1} \frac{1}{z^2} \frac{1}{z^2+10/3z+1} \; dz \\ =\frac{1}{6} \int_{|z|=1} \frac{1}{z^2} \sum_{q\ge 0} (-1)^q z^q (z+10/3)^q \; dz.$$
The only contribution in the series is from $q=1$ and it is $$-\frac{1}{6} 10/3 = -\frac{5}{9}.$$
Collecting everything we get $$-\frac{1}{i} \times 2\pi i\times \left(\frac{4}{9}-\frac{5}{9}\right) = \frac{2\pi}{9}.$$
Marko Riedel
- 61,317
-
@cambelot this, by the way, is a fairly standard way to use the residue theorem. Whenever you're taking an integral of a function over its period, this substitution lets you recast the problem as a residue calculation. – Ben Grossmann May 10 '15 at 04:08
1
HINT: use that $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ $$dx=\frac{2dt}{1+t^2}$$
Dr. Sonnhard Graubner
- 95,283
0
Just like the answer above that star with "Suppose we seek to evaluate ..."
Then, at the pole z=0, derivate once, in order that the residue evaluation do not colapse at the limit z=0.
This procedure is explained in "Mathematical Methods in the Physical Sciences", by Mary Boas. Chapter Functions of a complex Variable, section: 'evaluation of definite integral by use of the residue theorem'.
