I=∫(tanx+cotx)dx tanx=t2=∫(t+1t)2t1+t4dt=∫2(1+t2)1+t4dt sec2xdx=2tdt=2∫1+1t2(t−1t )2+2=2∫dzz2+2 t−1t=z=12tan−1(z2)+c=12tan−1(tanx−cotx2)+c I=∫sinx+cosxsinxcosx=2∫sinx+cosx2sinxcosxdx =2∫sinx+cosx1−(sinx−cosx)2dx=2 ∫ dt1−t2 =2sin−1(sinx−cosx)+c

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

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$\sqrt{2} \tan^{- 1} (\frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}}) + c$

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

$\sqrt{2} \tan^{- 1} (\frac{\sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{2}}) + c$

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

answer is A, C.

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Detailed Solution

$I = \int (\sqrt{\tan x} + \sqrt{\cot x}) d x \tan x = t^{2}$

$= \int (t + \frac{1}{t}) \frac{2 t}{1 + t^{4}} d t = \int \frac{2 (1 + t^{2})}{1 + t^{4}} d t \sec^{2} x d x = 2 t d t$

$= 2 \int \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t^{}})}^{2} + 2} = 2 \int \frac{d z}{z^{2} + 2} t - \frac{1}{t} = z$

$= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{z}{\sqrt{2}}) + c = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}}) + c I = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} d x = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - {(\sin x - \cos x)}^{2}}} d x = \sqrt{2} \int_{}^{} \frac{d t}{\sqrt{1 - t^{2}}} = \sqrt{2} \sin^{- 1} (\sin x - \cos x) + c$