integration of trigonometry functions

Question

I have question regarding the following two integrals;

$\int_0^{2\pi} \frac{\sin\theta}{A-\sin\theta} d\theta$, and $\int_0^{2\pi} \frac{1}{A-\sin\theta} d\theta$ where $A>0$. What method can be used to integrate them.

Answer 1

Hint:

Consider the substitution $$u=\sin \theta$$ and the identity $$\cos\theta =\pm\sqrt{1-\sin^2\theta}$$

Answer 2

Hint: substitute $$\sin(\theta)=\frac{2t}{1+t^2}$$ and $$d\theta=\frac{2}{1+t^2}dt$$ Your new integral is given by $$\int 4\,{\frac {t}{ \left( A{t}^{2}+A-2\,t \right) \left( {t}^{2}+1 \right) }} dt$$ to integrate use that $$\frac{4t}{(At^2-2t+A)(t^2+1)}=4+1/5\,{\frac {1}{ \left( t-2 \right) A}}+1/5\,{\frac {-t-2}{ \left( { t}^{2}+1 \right) A}} $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{2\pi}{\sin\pars{\theta} \over A - \sin\pars{\theta}}\,\dd \theta = -2\pi + A\int_{0}^{2\pi}{\dd\theta \over A - \sin\pars{\theta}}}$. I'll assume $\ds{A \in \mathbb{R}\setminus\bracks{-1,1}}$.

\begin{align} \int_{0}^{2\pi}{\dd\theta \over A - \sin\pars{\theta}} & = \int_{-\pi}^{\pi}{\dd\theta \over A + \sin\pars{\theta}} = \int_{0}^{\pi}\bracks{{1 \over A + \sin\pars{\theta}} + {1 \over A - \sin\pars{\theta}}}\dd\theta \\[5mm] & = 2A\int_{0}^{\pi}{\dd\theta \over A^{2} - \sin^{2}\pars{\theta}} = 2A\int_{-\pi/2}^{\pi/2}{\dd\theta \over A^{2} - \cos^{2}\pars{\theta}} = 4A\int_{0}^{\pi/2}{\dd\theta \over A^{2} - \cos^{2}\pars{\theta}} \\[5mm] & = 4A\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over A^{2}\sec^{2}\pars{\theta} - 1} \,\dd\theta = 4A\int_{0}^{\pi/2}{\sec^{2}\pars{\theta} \over A^{2}\tan^{2}\pars{\theta} + A^{2} - 1}\,\dd\theta \\[5mm] & = 4A\,{1 \over A^{2} - 1}\,{\root{A^{2} - 1} \over \verts{A}}\int_{0}^{\pi/2} {\verts{A}\sec^{2}\pars{\theta}/\root{A^{2} - 1} \over \bracks{\verts{A}\tan\pars{\theta}/\root{A^{2} - 1}}^{2} + 1}\,\dd\theta \\[5mm] & = {4\,\mrm{sgn}\pars{A} \over \root{A^{2} - 1}}\ \underbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}_{\ds{=\ {\pi \over 2}}}\ =\ \bbx{2\pi\,{\mrm{sgn}\pars{A} \over \root{A^{2} - 1}}} \end{align}

where $\ds{t = {\verts{A} \over \root{A^{2} - 1}}\,\tan\pars{\theta}}$.