If I=∫(cot⁡x−tan⁡x)dx, then I equals

I=∫cos⁡x−sin⁡xcos⁡xsin⁡xdrPut sin x + cos x = t, so that 2sin x cos x = t2 - 1∴I=2∫dtt2−1=2logt+t2−1+c=2log⁡|sin⁡x+cos⁡x+sin⁡2x|+c

If I=∫(cot⁡x−tan⁡x)dx, then I equals

I=∫cos⁡x−sin⁡xcos⁡xsin⁡xdrPut sin x + cos x = t, so that 2sin x cos x = t2 - 1∴I=2∫dtt2−1=2logt+t2−1+c=2log⁡|sin⁡x+cos⁡x+sin⁡2x|+c

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If $I = \int (\sqrt{\cot x} - \sqrt{\tan x}) dx$ , then I equals

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$\sqrt{2} \log (\sqrt{\tan x} - \sqrt{\cot x}) + C$

$\sqrt{2} \log | \sin x + \cos x + \sqrt{\sin 2 x} | + C$

$\sqrt{2} \log | \sin x - \cos x + \sqrt{2} \sin xcos x | + C$

$\sqrt{2} \log | \sin (x + π / 4) + \sqrt{2} \sin xcos x | + C$

answer is B.

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Detailed Solution

$I = \int \frac{\cos x - \sin x}{\sqrt{\cos xsin x}} dr$
Put sin x + cos x = t, so that 2sin x cos x = t² - 1
$\begin{array}{l} ∴ I = \sqrt{2} \int \frac{dt}{\sqrt{t^{2} - 1}} = \sqrt{2 \log} |t + \sqrt{t^{2} - 1}| + c \\ = \sqrt{2} \log | \sin x + \cos x + \sqrt{\sin 2 x} | + c \end{array}$