A) \[\pi \tan \frac{\pi }{8}\]
B) \[\log \tan \frac{\pi }{8}\]
C) \[\tan \frac{\pi }{8}\]
D) None of these
Correct Answer: A
Solution :
\[I=\int_{\pi /4}^{3\pi /4}{\frac{\varphi }{1+\sin \varphi }d\varphi }=\int_{\pi /4}^{3\pi /4}{\frac{\pi -\varphi }{1+\sin (\pi -\varphi )}d\varphi }\] \[\left\{ \because \frac{\pi }{4}+\frac{3\pi }{4}=\pi \right\}\] Þ \[2I=\int_{\pi /4}^{3\pi /4}{\frac{\pi }{1+\sin \varphi }d\varphi }\] On simplification, we get \[I=\pi (\sqrt{2}-1)=\pi \tan \frac{\pi }{8}.\]
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