Definite Integration Question 113
Question: $ \int_0^{\pi /2}{\frac{d\theta }{1+\tan \theta }}= $
[Roorkee 1980; MP PET 1996; DCE 1999]
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_0^{\pi /2}{\frac{d\theta }{1+\tan \theta }=\int_0^{\pi /2}{\frac{d\theta }{1+\tan ( \frac{\pi }{2}-\theta )}}} $
$ =\int_0^{\pi /2}{\frac{d\theta }{1+\cot \theta }} $
On adding, $ 2I=\int_0^{\pi /2}{( \frac{1}{1+\tan \theta }+\frac{1}{1+\cot \theta } )d\theta } $
= $ \int_0^{\pi /2}{d\theta =[\theta ]_0^{\pi /2}=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}} $ .
BETA
