I=∫tan⁡x+cos⁡x2sin⁡xdx=∫sin⁡xcos⁡x+cos⁡xsin⁡xdx=∫sin⁡x+cos⁡xsin⁡xcos⁡xdx=∫2(sin⁡x+cos⁡x)2sin⁡xcos⁡xdx=2∫sin⁡x+cos⁡xsin⁡2xdx Put sin⁡x−cos⁡x=t(cos⁡x+sin⁡x)dx=dt=2∫dt1−t2∴(sin⁡x−cos⁡x)2=t21−sin⁡2x=t2sin⁡2x=1−t2=2sin−1⁡t+c=2sin−1⁡(sin⁡x−cos⁡x)+c

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$\int (\sqrt{\tan x} + \sqrt{\cot x}) d x =$

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$\sqrt{2} \sin^{- 1} (\sin x + \cos x) + c$

$\frac{- 1}{\sqrt{2}} \sin^{- 1} (\sin x - \cos x) + c$

$\sqrt{2} \sin^{- 1} (\sin x - \cos x) + c$

$\sqrt{2} \cos^{- 1} (\sin x + \cos x) + c$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int (\sqrt{\tan x} + \frac{\sqrt{\cos x}}{\sqrt{2} \sin x}) d x \\ = \int (\frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}) d x \\ = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x \\ = \int \frac{\sqrt{2} (\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} d x \\ = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{\sin 2 x}} d x \\ Put \sin x - \cos x = t \\ (\cos x + \sin x) d x = d t \end{array}$

$\begin{array}{l} = \sqrt{2} \int \frac{d t}{\sqrt{1 - t^{2}}} (∴ (\sin x - \cos x)^{2} = t^{2} 1 - \sin 2 x = t^{2} \sin 2 x = 1 - t^{2}) \\ = \sqrt{2} \sin^{- 1} t + c \\ = \sqrt{2} \sin^{- 1} (\sin x - \cos x) + c \end{array}$