A) \[\frac{\pi }{2}\log 2\]
B) \[\pi \log 2\]
C) \[-\frac{\pi }{2}\log 2\]
D) \[-\pi \log 2\]
Correct Answer: C
Solution :
Put \[x=\sin \theta ,\]we get \[\int_{0}^{1}{\frac{\log x}{\sqrt{1-{{x}^{2}}}}dx=\int_{0}^{\pi /2}{\frac{\log \sin \theta .\cos \theta }{\cos \theta }}}\,d\theta \] \[=\int_{0}^{\pi /2}{\,\log \sin \theta }\,d\theta =-\frac{\pi }{2}\log 2\].
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