∫1+sin⁡x4dx =∫sin2⁡x8+cos2⁡x8+2sin⁡x8⋅cos⁡x8dx=∫sin⁡x8+cos⁡x8dx=−cos⁡x/81/8+sin⁡x/81/8=8sin⁡x8−cos⁡x8+c

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$\int \sqrt{1 + \sin (\frac{x}{4})} d x =$

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$\frac{1}{8} (\sin \frac{x}{8} - \cos \frac{x}{8}) + c$

$8 (\cos \frac{x}{8} - \sin \frac{x}{8}) + c$

$(\sin \frac{x}{8} + \cos \frac{x}{8}) + C$

$8 (\sin \frac{x}{8} - \cos \frac{x}{8}) + c$

answer is A.

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Detailed Solution

$\int \sqrt{1 + \sin (\frac{x}{4}) d x} \begin{array}{l} = \int \sqrt{((\sin^{2} \frac{x}{8} + \cos^{2} \frac{x}{8}) + (2 \sin \frac{x}{8} \cdot \cos \frac{x}{8})]} d x \\ = \int (\sin \frac{x}{8} + \cos \frac{x}{8}) d x \\ = \frac{- \cos x / 8}{1 / 8} + \frac{\sin x / 8}{1 / 8} = 8 (\sin \frac{x}{8} - \cos \frac{x}{8}) + c \end{array}$