Question

${\int }_0^{2\pi }\log \left[\frac{a+b\sec x}{a-b\sec x}\right]dx$. Show that the value of integral is independent of a and b.

Answer 1

Let $I={\int }_0^{2\pi }\log \left[\frac{a+b\sec x}{a-b\sec x}\right]dx$

Let $f\left(x\right)=\log \left[\frac{a+b\sec x}{a-b\sec x}\right]$

$f(-x)=\log \left[\frac{a+b\sec (-x)}{a-b\sec (-x)}\right]=\log \left[\frac{a+b\sec x}{a-b\sec x}\right]$

$\Rightarrow f(-x)=f\left(x\right)$

$\therefore I=2{\int }_0^{\pi }\log \left[\frac{a+b\sec x}{a-b\sec x}\right]dx$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I=2{\int }_0^{\pi }\log \left[\frac{a-b\sec x}{a+b\sec x}\right]dx$

$=-2{\int }_0^{\pi }\log \left[\frac{a+b\sec x}{a-b\sec x}\right]dx$

$\Rightarrow I=-I$

$\Rightarrow 2I=0$

$\Rightarrow I=0$