MCQ
$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }} = } $
- A${\cos ^{ - 1}}(x - 1) + c$
- ✓${\sin ^{ - 1}}(x - 1) + c$
- C${\cos ^{ - 1}}(1 + x) + c$
- D${\sin ^{ - 1}}(1 - x) + c$
✓
Answer
Correct option: B.
${\sin ^{ - 1}}(x - 1) + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }}} = \int_{}^{} {\frac{{dx}}{{\sqrt {1 - {{(x - 1)}^2}} }}} = {\sin ^{ - 1}}(x - 1) + c.$
(b)$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }}} = \int_{}^{} {\frac{{dx}}{{\sqrt {1 - {{(x - 1)}^2}} }}} = {\sin ^{ - 1}}(x - 1) + c.$
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