A) \[\log \frac{4}{3}\]
B) \[\log \frac{1}{3}\]
C) \[\log \frac{3}{4}\]
D) None of these
Correct Answer: A
Solution :
Put \[\sin x=t\Rightarrow \cos x\,dx=dt,\] so that reduced integral is \[\int_{0}^{1}{\left( \frac{1}{1+t}-\frac{1}{2+t} \right)\,\,dt=[\log (1+t)-\log (2+t)]_{0}^{1}}\] \[=\log \frac{2}{3}-\log \frac{1}{2}=\log \frac{4}{3}\].
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