The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\ln (1 + \frac{\sin x}{2})}{\sin x} d x =$

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$\frac{π^{2}}{3}$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π^{2}}{6}$

answer is B.

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Detailed Solution

$f (b) = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\ln (1 + b \sin x)}{\sin x} d x, f' (b) = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{1 + b \sin x} d x, f (b) = π \sin^{- 1} (b)$

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