MCQ
$\int_0^{\pi /2} {\frac{{d\theta }}{{1 + \tan \theta }}} = $
- A$\pi $
- B$\frac{\pi }{2}$
- C$\frac{\pi }{3}$
- ✓$\frac{\pi }{4}$
✓
Answer
Correct option: D.
$\frac{\pi }{4}$
d
(d) $I = \int_0^{\pi /2} {\frac{{d\theta }}{{1 + \tan \theta }} = \int_0^{\pi /2} {\frac{{d\theta }}{{1 + \tan \left( {\frac{\pi }{2} - \theta } \right)}}} } $
(d) $I = \int_0^{\pi /2} {\frac{{d\theta }}{{1 + \tan \theta }} = \int_0^{\pi /2} {\frac{{d\theta }}{{1 + \tan \left( {\frac{\pi }{2} - \theta } \right)}}} } $
$ = \int_0^{\pi /2} {\frac{{d\theta }}{{1 + \cot \theta }}} $
On adding, $2I = \int_0^{\pi /2} {\left( {\frac{1}{{1 + \tan \theta }} + \frac{1}{{1 + \cot \theta }}} \right)\,d\theta } $
$= \int_0^{\pi /2} {d\theta = [\theta ]_0^{\pi /2} = \frac{\pi }{2} \Rightarrow I = \frac{\pi }{4}} $.
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