If x∈0, 1 then ∫cos2cot-11-x1+xdx=

I=∫cos2cot-11-x1+xdxPut x=cos2θ=cos2cot-11-cos2θ1+cos2θ =cos2cot-12sin2θ2cos2θ =cos2cot-1 tanθ =cos2cot-1cotπ2-θ =cos2π2-θ =cosπ-2θ =-cos2θ =-x I= ∫ -x dx=-x22+C

If x∈0, 1 then ∫cos2cot-11-x1+xdx=

I=∫cos2cot-11-x1+xdxPut x=cos2θ=cos2cot-11-cos2θ1+cos2θ =cos2cot-12sin2θ2cos2θ =cos2cot-1 tanθ =cos2cot-1cotπ2-θ =cos2π2-θ =cosπ-2θ =-cos2θ =-x I= ∫ -x dx=-x22+C

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If $x \in (0, 1)$ then $\int \cos \{2 c o t^{- 1} \sqrt{\frac{1 - x}{1 + x}}\} d x =$

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$- c o t^{- 1} x + c$

$\frac{- x^{2}}{2} + c$

$c o t^{- 1} x + c$

$\frac{x^{2}}{2} + c$

answer is D.

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Detailed Solution

$I = \int \cos \{2 c o t^{- 1} \sqrt{\frac{1 - x}{1 + x}}\} d x$

$P u t x = \cos 2 θ$

$= \cos \{2 c o t^{- 1} \sqrt{\frac{1 - \cos 2 θ}{1 + \cos 2 θ}}\} = \cos \{2 c o t^{- 1} \sqrt{\frac{2 \sin^{2} θ}{2 \cos^{2} θ}}\} = \cos \{2 c o t^{- 1} \tan θ\} = \cos \{2 c o t^{- 1} (c o t (\frac{π}{2} - θ))\} = \cos \{2 (\frac{π}{2} - θ)\} = \cos \{π - 2 θ\} = - \cos 2 θ = - x I = \int - x d x = \frac{- x^{2}}{2} + C$