$\int \sqrt{1 + cos e c x} d x$ is equal to

{sin}^{- 1} (2 sin x + 1) + c

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- {sin}^{- 1} (2 sin x + 1) + c

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{sin}^{- 1} (2 sin x - 1) + c

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- {sin}^{- 1} (2 sin x - 1) + c

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Solution

The correct option is C ${sin}^{- 1} (2 sin x - 1) + c$
Let, $\sqrt{1 + c o s e c x} = t => \frac{- c o s e c x c o t x}{2 \sqrt{1 + c o s e c x}} d x = d t s i n x$
$= \frac{1}{(t^{2} - 1)}$ and $c o s x = \sqrt{1 - (\frac{1}{(t^{2} - 1)})^{2}} = \frac{t \sqrt{t^{2} - 2}}{t^{2} - 1}$
Substituting back,
$= \int \frac{- 2 t d t}{(t^{2} - 1) \sqrt{t^{2} - 2}}$
Let, $t^{2} - 2 = m => 2 t d t = d m$
Substituting back,
$\int \frac{- d m}{(m + 1) \sqrt{m}}$
Let, $m = p^{2} => d m = 2 p d p$
Substituting back,
$= \int \frac{- 2 p d p}{(p^{2} + 1) p}$
$= \int \frac{- 2 d p}{(p^{2} + 1)}$
$= - 2 t a n^{- 1} p + c = - 2 s i n^{- 1} \frac{p}{\sqrt{p^{2} + 1}} + c$
$= - 2 s i n^{- 1} \sqrt{\frac{t^{2} - 2}{t^{2} - 1}} + c$
$= - 2 s i n^{- 1} \sqrt{1 - s i n x} + c$

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