If∫cosec⁡x+1dx=kfog(x)+c, where k is real constant, then

I=∫cosec x+1dx=∫cot xcosec x−1dxPut corsec x−1=t2 or −cosec x cot x dx=2tdtI=−∫−cot x cosec xcosec xcosec x−1dx=−∫2dt1+t2=−2 tan−1 t+c=−2tan−1cosec x−1+C=−2π2−cot−1cosec x−1+C=2 cot−1cosec x−1+C=2cot−1cot xcosec x+1+C

If∫cosec⁡x+1dx=kfog(x)+c, where k is real constant, then

I=∫cosec x+1dx=∫cot xcosec x−1dxPut corsec x−1=t2 or −cosec x cot x dx=2tdtI=−∫−cot x cosec xcosec xcosec x−1dx=−∫2dt1+t2=−2 tan−1 t+c=−2tan−1cosec x−1+C=−2π2−cot−1cosec x−1+C=2 cot−1cosec x−1+C=2cot−1cot xcosec x+1+C

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If $\int \sqrt{cosec x + 1} d x = k f o g (x) + c,$ where k is real constant, then

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$k = - 2, f (x) = \cot^{- 1} x, g (x) = \sqrt{cosec x - 1}$

$k = - 2, f (x) = \tan^{- 1} x, g (x) = \sqrt{cosec x - 1}$

$k = 2, f (x) = \tan^{- 1} x, g (x) = \frac{\cot x}{\sqrt{cosec x - 1}}$

$k = 2, f (x) = \cot^{- 1} x, g (x) = \frac{\cot x}{\sqrt{cosec x + 1}}$

answer is B, D.

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Detailed Solution

$I = \int \sqrt{cosec x + 1} d x = \int \frac{\cot x}{\sqrt{cosec x - 1}} d x$

Put corsec $x - 1 = t^{2} or - cosec x \cot x d x = 2 t d t$

$\begin{aligned} I = - \int \frac{- \cot x cosec x}{cosec x \sqrt{cosec x - 1}} d x \\ = - \int \frac{2 d t}{1 + t^{2}} = - 2 \tan^{- 1} t + c \\ = - 2 \tan^{- 1} \sqrt{cosec x - 1} + C \\ = - 2 [\frac{π}{2} - \cot^{- 1} \sqrt{cosec x - 1}] + C \\ = 2 \cot^{- 1} \sqrt{cosec x - 1} + C \\ = 2 \cot^{- 1} \frac{\cot x}{\sqrt{cosec x + 1}} + C \end{aligned}$