MCQ
$\int_0^{\pi /2} {\frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}\,dx = } $
- A$0$
- B$\frac{\pi }{2}$
- ✓$\frac{\pi }{4}$
- DNone of these
and $I = \int_0^{\pi /2} {\frac{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}dx} $
$I = \int_0^{\pi /2} {\frac{{\sqrt {\sin x} }}{{\sqrt {\cos x + } \sqrt {\sin x} }}} \,dx$....$(ii)$
Adding $(i)$ and $(ii),$ we get
$2I = \int_0^{\pi /2} {(1)dx = \frac{\pi }{2} \Rightarrow I = \frac{\pi }{4}} $.
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$\left| {\begin{array}{*{20}{c}}
{ - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\
{\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\
{ - 1 + \cos B}&{ - 1 + \cos A}&{ - 1}
\end{array}} \right|$
$f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+f\left(\frac{3}{20}\right)+\ldots \ldots+f\left(\frac{39}{20}\right)$ is equal to ....... .
$\frac{19}{8}$
$\frac{8}{19}$
$\frac{19}{2}$
$\frac{3}{4}$