_ ^ 1 ^ - 1 x. 1 - x^2 dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{{{\cos }^{ - 1}}x.\sqrt {1 - {x^2}} }}dx = } $

A
$\log ({\cos ^{ - 1}}x) + c$
✓
$ - \log ({\cos ^{ - 1}}x) + c$
C
$ - \frac{1}{{2{{({{\cos }^{ - 1}}x)}^2}}} + c$
D
None of these

✓

Answer

Correct option: B.

$ - \log ({\cos ^{ - 1}}x) + c$

b
(b) 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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