A) \[\log ({{\cos }^{-1}}x)+c\]
B) \[-\log ({{\cos }^{-1}}x)+c\]
C) \[-\frac{1}{2{{({{\cos }^{-1}}x)}^{2}}}+c\]
D) None of these
Correct Answer: B
Solution :
Put \[{{\cos }^{-1}}x=t\Rightarrow -\frac{1}{\sqrt{1-{{x}^{2}}}}\,dx=dt,\] then \[\int_{{}}^{{}}{\frac{1}{{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}}\,dx=-\int_{{}}^{{}}{\frac{1}{t}\,dt}}=-\log t+c=\log \frac{1}{t}+c\] \[=-\log ({{\cos }^{-1}}x)+c.\]
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