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

MCQ

$\int_{}^{} {\frac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\;} dx = $

✓
$x - \sqrt {1 - {x^2}} {\sin ^{ - 1}}x + c$
B
$x + \sqrt {1 - {x^2}} {\sin ^{ - 1}}x + c$
C
$\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - x + c$
D
None of these

✓

Answer

Correct option: A.

$x - \sqrt {1 - {x^2}} {\sin ^{ - 1}}x + c$

a
(a) Putting ${\sin ^{ - 1}}x = t \Rightarrow \frac{1}{{\sqrt {1 - {x^2}} }}\,dx = dt,$ we get
$\int_{}^{} {\frac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\,dx = } \int_{}^{} {t\sin t\,dt = - t\cos t + \sin t + c} $
$ = - {\sin ^{ - 1}}x\cos ({\sin ^{ - 1}}x) + \sin ({\sin ^{ - 1}}x) + c$
$ = x - {\sin ^{ - 1}}x\sqrt {1 - {x^2}} + c.$

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