Complex Integration Problem. Please help.

Question

Please help me with this one.

Calculate the integral:

$$\int_0^{2\pi} \frac{\mathrm{d}t}{a\cos t+b\sin t+c}$$ as $\sqrt{a^2+b^2}=1<c$.

I'm working on it for quite a while and somehow I can't manage to solve this problem.

Answer 1

First use trigonometric identity: $$a\cos(x)+b\sin(x)=\sqrt{a^2+b^2}\cos(x-\text{atan2}(b,a)) = \cos(x-\text{atan2}(b,a))$$

then integral is:

$$I = \int_0^{2\pi}\frac{1}{\cos(t-\text{atan2}(b,a))+c}\,dt$$

because we are integrating over the full period of cosine, phase term can be dropped.

$$I = \int_0^{2\pi}\frac{1}{\cos(t)+c}\,dt$$

Now if you use weierstrass substitution you get the integral:

$$I = \int_{-\infty}^{\infty} = \frac{2}{1+c+(c-1)t^2}\,dt$$

This one should be easier to solve.

Answer 2

I suggest you use the tangent half-angle substitution (also called the Weierstrass substitution). Your problem becomes quite simple (almost when the conditions are taken into account). If you need more help, just tell.

Answer 3

As the problem is tagged complex analysis:

Using $z=e^{it} \, 0 \leq t \leq 2\pi$ we get

$$\int_0^{2\pi} \frac{\mathrm{d}t}{a\cos t+b\sin t+c}=\int_{|z|=1} \frac{1}{a\frac{z+\frac{1}{z}}{2}+b\frac{z-\frac{1}{z}}{2i}+c} \frac{dz}{iz}\\=\frac1i\int_{|z|=1} \frac{2i}{ai(z^2+1)+b(z^2-1)+2icz}dz\\=2\int_{|z|=1} \frac{1}{(ai+b)z^2+2icz+(ai-b)}dz$$

Solving

$$(ai+b)z^2+2icz+(ai-b)=0$$ you get $$z_{1,2}= \frac{-2ci\pm \sqrt{-4c^2+4(a^2+b^2)}}{2ai+2b}=i\frac{-c\pm \sqrt{c^2-(a^2+b^2)}}{ai+b}$$

Now, check if the poles are inside the circle $|z|=1$ and calculate the residue at each pole (which is easy since the poles are simple).

Answer 4

$\sqrt{a^2+b^2}=1\iff a^2+b^2=1\iff\begin{align}&a=\cos\alpha\\&b=-\sin\alpha\end{align}\iff\displaystyle\int_0^{2\pi}\frac{dt}{a\cos t+b\sin t+c}=$ $=\displaystyle\int_0^{2\pi}\frac{dt}{\cos\alpha\cos t-\sin\alpha\sin t+c}=\displaystyle\int_0^{2\pi}\frac{dt}{\cos(\alpha+t)+c}=\displaystyle\int_0^{2\pi}\frac{dt}{c+\cos t}$ . Now use the Weierstrass substitution.

Answer 5

Yes, it's quite the way I did it:

Like @0912 wrote I got this integrand: $\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

$$\oint_0^{2\pi}\frac{\mathrm dt}{\sin t+c}\xRightarrow[\text{Substitution}]{}\left[z=e^{it},\mathrm dt=\frac{\mathrm dz}{iz}\right]\xRightarrow[\text{After some calculations}]{}2\oint_{D(0,1)}\frac{\mathrm dz}{z^2+i2cz-}$$

Now, using this substitution we get to calculate an integral along a curve (the unit circle):

$$\begin{align}&=2\oint_{D(0,1)}\frac{\mathrm dz}{(z+ic-\sqrt{1-c^2})(z+ic+\sqrt{1-c^2})}\\ \, &\\ \\&=\frac{1}{2\pi i}\oint_{D(0,1)}\frac{4\pi i\cdot \mathrm dz}{(z+ic-\sqrt{1-c^2})(z+ic+\sqrt{1- c^2})}\end{align}$$

Using Cauchy Intergal Formula:

$$=\frac{4\pi i}{\sqrt{1-c^2}-ic+ic+\sqrt{1-c^2}}=\frac{4\pi i}{2\sqrt{1-c^2}}=\frac{2\pi i}{\sqrt{1-c^2}}$$