∫01dxx2+2xsinα+1=∫01dxx+sinα2+1-sin2α =∫01dxx+sinα2+cos2α =1cosαtan-1x+sinαcosα01 (∴ ∫1x2+a2dx=1atan−1xa) =secαtan-11+sinαcosα-tan-1tanα =secαtan-1tanπ4+α2-α =secαπ4+α2-α =secαπ4-α2 ∴ 1+sinαcosα=cos2α2+sin2α2+2sinα2cosα2cosα=(cosα2+sinα2)2cos2α2+sin2α2=cosα2+sinα2cosα2−sinα2=1+tanα21−tanα2=tan(π4+α2)

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$\int_{0}^{1} \frac{d x}{x^{2} + 2 x \sin α + 1} =$

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$(\frac{π}{4} - \frac{α}{2}) \cos e c α$

$(\frac{π}{4} + \frac{α}{2}) s e c α$

$(\frac{π}{4} + \frac{α}{2}) \cos e c α$

$(\frac{π}{4} - \frac{α}{2}) s e c α$

answer is D.

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Detailed Solution

$\int_{0}^{1} \frac{d x}{x^{2} + 2 x \sin α + 1}$

$= \int_{0}^{1} \frac{d x}{{(x + \sin α)}^{2} + 1 - \sin^{2} α} = \int_{0}^{1} \frac{d x}{{(x + \sin α)}^{2} + \cos^{2} α} = {(\frac{1}{\cos α} \tan^{- 1} (\frac{x + \sin α}{\cos α}))}_{0}^{1} (∴ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a}) = s e c α [\tan^{- 1} (\frac{1 + \sin α}{\cos α}) - \tan^{- 1} (\tan α)] = s e c α (\tan^{- 1} (\tan (\frac{π}{4} + \frac{α}{2})) - α) = s e c α (\frac{π}{4} + \frac{α}{2} - α) = s e c α (\frac{π}{4} - \frac{α}{2})$

$\begin{array}{l} (∴ \frac{1 + \sin α}{\cos α} = \frac{\cos^{2} \frac{α}{2} + \sin^{2} \frac{α}{2} + 2 \sin \frac{α}{2} \cos \frac{α}{2}}{\cos α}) \\ = \frac{{(\cos \frac{α}{2} + \sin \frac{α}{2})}^{2}}{\cos^{2} \frac{α}{2} + \sin^{2} \frac{α}{2}} = \frac{\cos \frac{α}{2} + \sin \frac{α}{2}}{\cos \frac{α}{2} - \sin \frac{α}{2}} \\ = \frac{1 + \tan \frac{α}{2}}{1 - \tan \frac{α}{2}} = \tan (\frac{π}{4} + \frac{α}{2}) \end{array}$