Question
The value of $$\int_0^{\frac{\pi }{4}} {\log \left( {1 + \tan \,x} \right)dx} $$ is equal to :
A.
$$\frac{\pi }{8}{\log _e}2$$
B.
$$\frac{\pi }{4}{\log _e}2$$
C.
$$\frac{\pi }{4}$$
D.
none of these
Answer :
$$\frac{\pi }{8}{\log _e}2$$
Solution :
$$\eqalign{
& I = \int_0^{\frac{\pi }{4}} {\log } \left( {1 + \tan \,x} \right)dx \cr
& \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log \left\{ {1 + \tan \left( {\frac{\pi }{4} - x} \right)} \right\}dx} \cr
& \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log } \left\{ {1 + \frac{{1 - \tan \,x}}{{1 + \tan \,x}}} \right\}dx \cr
& \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log } \frac{2}{{1 + \tan \,x}}dx \cr
& \,\,\,\,\, = \int_0^{\frac{\pi }{4}} {\log } \,2\,dx - I \cr
& \therefore I = \frac{1}{2}\int_0^{\frac{\pi }{4}} {\log } \,2\,dx = \frac{1}{2}.\log \,2.\frac{\pi }{4} \cr} $$