Definite Integration Question 480
Question: The value of $ \int _{\pi /4}^{3\pi /4}{\frac{\varphi }{1+\sin \varphi }d\varphi ,} $ is
[AI CBSE 1990; IIT 1993]
Options:
A) $ \pi \tan \frac{\pi }{8} $
B) $ \log \tan \frac{\pi }{8} $
C) $ \tan \frac{\pi }{8} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int _{\pi /4}^{3\pi /4}{\frac{\varphi }{1+\sin \varphi }d\varphi }=\int _{\pi /4}^{3\pi /4}{\frac{\pi -\varphi }{1+\sin (\pi -\varphi )}d\varphi } $
$ { \because \frac{\pi }{4}+\frac{3\pi }{4}=\pi } $
therefore $ 2I=\int _{\pi /4}^{3\pi /4}{\frac{\pi }{1+\sin \varphi }d\varphi } $
On simplification, we get $ I=\pi (\sqrt{2}-1)=\pi \tan \frac{\pi }{8}. $
BETA
