Question
Solve this : $$\int\limits_0^{2\pi } {\log \left( {\frac{{a + b\,\sec \,x}}{{a - b\,\sec \,x}}} \right)dx = ?} $$
A.
$$0$$
B.
$$\frac{\pi }{2}$$
C.
$$\frac{{\pi \left( {a + b} \right)}}{{a - b}}$$
D.
$$\frac{\pi }{2}\left( {{a^2} - {b^2}} \right)$$
Answer :
$$0$$
Solution :
$$\eqalign{
& \int\limits_0^{2\pi } {\log \left( {\frac{{a + b\,\sec \,x}}{{a - b\,\sec \,x}}} \right)dx} \cr
& = 2\int\limits_0^\pi {\log \left( {\frac{{a + b\,\sec \,x}}{{a - b\,\sec \,x}}} \right)dx} \cr
& = 2\int\limits_0^\pi {\log \left( {a + b\,\sec \,x} \right)dx} - 2\int\limits_0^\pi {\log \left( {a - b\,\sec \left( {\pi - x} \right)} \right)dx} \cr
& = 2\int\limits_0^\pi {\log \left( {a + b\,\sec \,x} \right)dx} - 2\int\limits_0^\pi {\log \left( {a + b\,\sec \,x} \right)dx} \cr
& = 0 \cr} $$