doubt reposting question of integration

Question

$\int_{0}^{\frac{\pi }{2}} \frac{cos x}{2\sqrt{1 - sin x}} dx$

$cos 2x = 2 cos^2 x -1$

so , cos x = $cos x = 2 cos^2 (\frac{x}{2}) -1$

1 - sin x = (sin x/2 - cos x/2)$^{2}$

$\frac{(2 cos^2\frac{x}{2} - 1)}{2(sin\frac{x}{2} - cos \frac{x}{2})}$

$\frac{(2 cos^2\frac{x}{2} - (sin^2\frac{x}{2}+cos^2\frac{x}{2}))}{2(sin\frac{x}{2} - cos \frac{x}{2})}$

$\frac{ cos^2\frac{x}{2} - (sin^2\frac{x}{2})}{2(sin\frac{x}{2} - cos \frac{x}{2})}$

=$\frac{ (cos\frac{x}{2} + sin\frac{x}{2})(cos\frac{x}{2}-sin\frac{x}{2})}{2(sin\frac{x}{2} - cos \frac{x}{2})}$

= $\frac{-1}{2} \int_{0}^{\frac{\pi }{2}} (cos \frac{x}{2} + sin\frac{x}{2}) dx$

= $\frac{-1}{2} (2 sin \frac{x}{2} - 2 cos \frac{x}{2})$  

range from 0  to $\frac{\pi }{2}$

$\frac{-1}{2}((\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{2}} ) - ( 0 - 2)) = -1$

now , conclusion is  dont do like this because  1- sinx =$(sin\frac{x}{2} - cos\frac{x}{2})^2 = ( cos\frac{x}{2} - sin\frac{x}{2})^2$

if i would have taken  1-sinx =  $( cos\frac{x}{2} - sin\frac{x}{2})$

then answer i get = 1   @check it  i think iam making mistake answer should be 1 in both  cases  , what is reason behind it  square is always positive so we should get +1 answer ??

Comments

@sumit goyal 1

$(1-sinx)=(sin\frac{x}{2}-cos\frac{x}{2})^2$

$\Rightarrow sin\frac{x}{2}-cos\frac{x}{2}=\pm \sqrt{1-sinx}$...........(i), till now everything is trivial.

now see,

in interval $(0,\frac{\pi }{2})$ , $cos\frac{x}{2}>sin\frac{x}{2}$     $\therefore sin\frac{x}{2}-cos\frac{x}{2}<0(-ve)$

therefore in equation (i) RHS should be negative, and hence

$sin\frac{x}{2}-cos\frac{x}{2}=-\sqrt{1-sinx}$    or     $\sqrt{1-sinx}=cos\frac{x}{2}-sin\frac{x}{2}$

therefore for an interval $(0,\frac{\pi }{2})$,  $\sqrt{1-sinx}=cos\frac{x}{2}-sin\frac{x}{2}$ is correct.

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@joshi_nitish good hmm i  thought something like this but didnot try to check this good one thanku  , its better  that we should g for substitution