_0^ /4 (1 + ) ,d = - Vidyadip

MCQ

$\int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta = } $

A
$\frac{\pi }{4}\log 2$
B
$\frac{\pi }{4}\log \frac{1}{2}$
✓
$\frac{\pi }{8}\log 2$
D
$\frac{\pi }{8}\log \frac{1}{2}$

✓

Answer

Correct option: C.

$\frac{\pi }{8}\log 2$

c
(c) $I = \int_0^{\pi /4} {\,\,\,\log (1 + \tan \theta )d\theta } $

==> $I = \int_0^{\pi /4} {\log \left\{ {1 + \tan \left( {\frac{\pi }{4} - \theta } \right)} \right\}} \,d\theta $

==> I = $\int_0^{\pi /4} {\log \left( {1 + \frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)\,d\theta } $

==> I = $\int_0^{\pi /4} {\log 2d\theta - \int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta } } $

$ \Rightarrow I = \frac{1}{2}\int_0^{\pi /4} {\log 2d\theta = \frac{{\log 2}}{2}|\theta |_0^{\pi /4} = \frac{\pi }{8}\log 2} $.

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