Question

Evaluate the following integrals:

${\int }_{-a}^a\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right)d\theta$

Answer 1

Let I = ${\int }_{-a}^a\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right)d\theta$

Consider $f\left(\theta \right)=\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right)$.

$$\begin{gathered} f\left(-\theta \right)=\log \left(\frac{a-\sin \left(-\theta \right)}{a+\sin \left(-\theta \right)}\right) \\ =\log \left(\frac{a+\sin \theta }{a-\sin \theta }\right)\left[\sin \left(-x\right)=-\sin x\right] \\ =\log {\left(\frac{a-\sin \theta }{a+\sin \theta }\right)}^{-1} \\ =-\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right)\left[\log a^b=b\log a\right] \\ =-f\left(\theta \right) \end{gathered}$$

$$\begin{gathered} \therefore f\left(-\theta \right)=-f\left(\theta \right) \\ \Rightarrow I={\int }_{-a}^a\log \left(\frac{a-\sin \theta }{a+\sin \theta }\right)d\theta =0\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right] \end{gathered}$$