∫1−sin⁡xcos⁡xdx=12f(x)+C then f(0)=

I=∫1−sin⁡xcos⁡x_dx=I=∫cos⁡x2−sin⁡x2cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2dxI=∫1212cos⁡x2+12sin⁡x2dx=12∫1sin⁡π4+x2dx=12∫cos⁡ecx2+π4dx=12log⁡cos⁡ecx2+π4−cot⁡x2+π4+Cf(x)=log⁡cos⁡ecx2+π4−cot⁡x2+π4f0=log2−1

∫1−sin⁡xcos⁡xdx=12f(x)+C then f(0)=

I=∫1−sin⁡xcos⁡x_dx=I=∫cos⁡x2−sin⁡x2cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2dxI=∫1212cos⁡x2+12sin⁡x2dx=12∫1sin⁡π4+x2dx=12∫cos⁡ecx2+π4dx=12log⁡cos⁡ecx2+π4−cot⁡x2+π4+Cf(x)=log⁡cos⁡ecx2+π4−cot⁡x2+π4f0=log2−1

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$\int \frac{\sqrt{1 - \sin x}}{\cos x} d x = \frac{1}{\sqrt{2}} f (x) + C then f (0) =$

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$\log (\sqrt{2} - 1)$

$\frac{1}{\sqrt{2}} \log (\sqrt{2} - 1)$

$\frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

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answer is A.

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Detailed Solution

$\begin{array}{l} \underline{I = \int \frac{\sqrt{1 - \sin x}}{\cos x}} d x = \\ I = \int \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2}) (\cos \frac{x}{2} + \sin \frac{x}{2})} d x \\ I = \int \frac{1}{\sqrt{2} (\frac{1}{\sqrt{2}} \cos \frac{x}{2} + \frac{1}{\sqrt{2}} \sin \frac{x}{2})} d x \\ = \frac{1}{\sqrt{2}} \int \frac{1}{\sin (\frac{π}{4} + \frac{x}{2})} d x \\ = \frac{1}{\sqrt{2}} \int \cos e c (\frac{x}{2} + \frac{π}{4}) d x \\ = \frac{1}{\sqrt{2}} \log (\cos e c (\frac{x}{2} + \frac{π}{4}) - \cot (\frac{x}{2} + \frac{π}{4})) + C \\ f (x) = \log (\cos e c (\frac{x}{2} + \frac{π}{4}) - \cot (\frac{x}{2} + \frac{π}{4})) \end{array}$