MCQ
$\int {\,\,\sqrt {1\,\, + \,\,\csc \,x} } dx$ equals
- A$2\, \sin^{ -1} \sqrt {\sin \,x} + c$
- B$\sqrt 2 cos^{ -1} \sqrt {\cos \,x} + c$
- C$cos ^{-1} (1 - 2\, \sin\, x) + c$
- ✓Both $(A)$ and $(C)$
✓
Answer
Correct option: D.
Both $(A)$ and $(C)$
d
$I =$ $\frac{{\sqrt {\left( {1\, + \,\sin \,x} \right)\,\,\left( {1\, - \,\sin \,x} \right)} }}{{\sqrt {\sin \,x\,\,\left( {1\, - \,\sin \,x} \right)} }}$
$I =$ $\frac{{\sqrt {\left( {1\, + \,\sin \,x} \right)\,\,\left( {1\, - \,\sin \,x} \right)} }}{{\sqrt {\sin \,x\,\,\left( {1\, - \,\sin \,x} \right)} }}$
$=$ $\frac{{\cos \,x}}{{\sqrt {\sin \,x\,\,\left( {1\, - \,\sin \,x} \right)} }}$
$=$ $\frac{{\cos \,x}}{{\sqrt {{\textstyle{1 \over 4}}\,\, - \,\,{{\left( {{\textstyle{1 \over 2}}\,\, - \,\,\sin \,x} \right)}^2}} }}$
$=$ $\int {\frac{{ - dt}}{{\sqrt {\,{{\left( {{\textstyle{1 \over 2}}\,} \right)}^2} - {t^2}} }}}$
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