$\int \sqrt{2 \cos 2 x + \sin x} d x \cos x is equal to$

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$\frac{(8 \sin x - 1)}{64} \sqrt{33 - {(8 \sin x + 1)}^{2}} + \frac{33}{64} \sin^{- 1} (\frac{8 \sin x - 1}{\sqrt{33}}) + C$

$\frac{8 \sin x - 1}{64} \sqrt{33 + {(8 \sin x - 1)}^{2}} + \frac{33}{64} \sin^{- 1} (\frac{8 \sin x - 1}{\sqrt{33}}) + C$

$\frac{8 \sin x - 1}{64} \sqrt{33 - {(8 \sin x - 1)}^{2}} + \frac{33}{64} \sin^{- 1} (\frac{8 \sin x - 1}{\sqrt{33}}) + C$

$\frac{8 \sin x - 1}{64} \sqrt{{(8 \sin x - 1)}^{2} - 33} + \frac{33}{64} \sin^{- 1} (\frac{8 \sin x - 1}{\sqrt{33}}) + C$

answer is C.

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Detailed Solution

$Let I = \int \sqrt{2 \cos 2 x + \sin x} \cos x d x$

$\begin{aligned} = \int \sqrt{2 (1 - 2 \sin^{2} x) + \sin x} \cos x d x \\ = \sqrt{- 4 \sin^{2} x + \sin x + 2} \cos x d x \end{aligned}$

$Put \sin x = t$

$\begin{array}{l} = \int \sqrt{- 4 t^{2} + t + 2} d t \\ = \int \sqrt{(- 4) [t^{2} - \frac{1}{4} t - \frac{2}{4}]} d t \\ = \int \sqrt{(- 4) [{(t - \frac{1}{8})}^{2} - {(\frac{1}{8})}^{2} - \frac{1}{2}]} d t \\ = \int \sqrt{(- 4) [{(t - \frac{1}{8})}^{2} - \frac{33}{64}]} d t \end{array}$

$= 2 \int \sqrt{[{(\frac{\sqrt{33}}{8})}^{2} - {(t - \frac{1}{8})}^{2}]} d t$

$= 2 [(\frac{t - \frac{1}{8}}{2}) \sqrt{{(\frac{\sqrt{33}}{8})}^{2} - {(t - \frac{1}{8})}^{2}} + \frac{{(\frac{\sqrt{33}}{8})}^{2}}{2} \sin^{- 1} (\frac{t - \frac{1}{8}}{\frac{\sqrt{33}}{8}})] + C$