A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /2}{\frac{d\theta }{1+\tan \theta }}=\int_{0}^{\pi /2}{\frac{d\theta }{1+\tan \left( \frac{\pi }{2}-\theta \right)}}\] \[=\int_{0}^{\pi /2}{\frac{d\theta }{1+\cot \theta }}\] On adding, we get \[2I=\int_{0}^{\pi /2}{\left( \frac{1}{1+\tan \theta }+\frac{1}{1+\cot \theta } \right)d\theta }\] \[\int_{0}^{\pi /2}{d\theta =[\theta ]_{0}^{\pi /2}}=\frac{\pi }{2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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