<s>[INST] Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1<c$. I am lead to believe I should be using curves but I really don't understand what curves to choose and how to properly use them. I would appreaciate some help on the issue. 
 [/INST] We have
$$a\sin(t)+b\cos(t) = \sqrt{a^2+b^2}\sin(t+\phi) = \sin(t+\phi)$$
Hence, we have
\begin{align}
I & = \int_0^{2\pi} \dfrac{dt}{\sin(t+\phi)+c} = \int_0^{2\pi} \dfrac{dt}{\sin(t)+c} = \dfrac1c \int_0^{2\pi} \dfrac{dt}{1+\dfrac{\sin(t)}c}\\
& = \dfrac1c \sum_{k=0}^{\infty}\left(-\dfrac1c \right)^k \int_0^{2\pi} \sin^{k}(t)dt = \dfrac1c \sum_{k=0}^{\infty} \left(\dfrac1c\right)^{2k} \int_0^{2\pi} \sin^{2k}(t)dt\\
& = \dfrac4c \sum_{k=0}^{\infty} \left(\dfrac1c\right)^{2k} \int_0^{\pi/2} \sin^{2k}(t)dt
\end{align}
From here, we have
$$\int_0^{\pi/2} \sin^{2k}(t)dt = \dfrac{\pi}{2^{2k+1}}\dbinom{2k}k$$
Hence,
\begin{align}
I & = \dfrac{2\pi}c \sum_{k=0}^{\infty} \left(\dfrac1{2c}\right)^{2k} \dbinom{2k}k = \dfrac{2\pi}c \cdot \dfrac1{\sqrt{1-1/c^2}} = \dfrac{2\pi}{\sqrt{c^2-1}}
\end{align}
</s>