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Question: \[\int_{}^{}{\sqrt{\frac{\mathbf{1}\mathbf{-}\mathbf{x}}{\mathbf{1 + x}}}\mathbf{dx =}}\]...

∫1−x1+xdx=\int_{}^{}{\sqrt{\frac{\mathbf{1}\mathbf{-}\mathbf{x}}{\mathbf{1 + x}}}\mathbf{dx =}}

A

sin⁡−1x−121−x2+C\sin^{- 1}x - \frac{1}{2}\sqrt{1 - x^{2}} + C

B

sin⁡−1x+121−x2+C\sin^{- 1}x + \frac{1}{2}\sqrt{1 - x^{2}} + C

C

sin⁡−1x−1−x2+C\sin^{- 1}x - \sqrt{1 - x^{2}} + C

D

sin⁡−1x+1−x2+C\sin^{- 1}x + \sqrt{1 - x^{2}} + C

Answer

sin⁡−1x+1−x2+C\sin^{- 1}x + \sqrt{1 - x^{2}} + C

Explanation

Solution

Putx=cos⁡2θx = \cos 2\theta, then θ=12cos⁡−1x\theta = \frac{1}{2}\cos^{- 1}x ⇒ dx=−2sin⁡2θdθdx = - 2\sin 2\theta d\theta

∴I=−2∫1−cos⁡2θ1+cos⁡2θ.sin⁡2θdθ\therefore I = - 2\int_{}^{}{\sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}}.\sin 2\theta d\theta} ⇒I=−2∫2sin⁡2θ/2cos⁡2θ.sin⁡2θ.dθ⇒I=−2∫tan⁡θ.2sin⁡θcos⁡θdθ\Rightarrow I = - 2\int_{}^{}\sqrt{2\sin^{2}\theta/2\cos^{2}\theta}.\sin 2\theta.d\theta \Rightarrow I = - 2\int_{}^{}{\tan\theta.2\sin\theta\cos\theta d\theta} ⇒I=−2.2∫sin⁡2θdθ\Rightarrow I = - 2.2\int_{}^{}{\sin^{2}\theta d\theta} ⇒I=−2∫(1−cos⁡2θ)dθ\Rightarrow I = - 2\int_{}^{}{(1 - \cos 2\theta)d\theta} ⇒I=−2[θ−sin⁡2θ2]+C1\Rightarrow I = - 2\left\lbrack \theta - \frac{\sin 2\theta}{2} \right\rbrack + C_{1}

⇒I=−2θ+sin⁡2θ+C1\Rightarrow I = - 2\theta + \sin 2\theta + C_{1} ⇒I=−cos⁡−1x+1−x2+C1⇒I=−π2+sin⁡−1x+1−x2+C1⇒I=sin⁡−1x+1−x2+C\Rightarrow I = - \cos^{- 1}x + \sqrt{1 - x^{2}} + C_{1} \Rightarrow I = - \frac{\pi}{2} + \sin^{- 1}x + \sqrt{1 - x^{2}} + C_{1} \Rightarrow I = \sin^{- 1}x + \sqrt{1 - x^{2}} + Cwhere C=C1−π2C = C_{1} - \frac{\pi}{2}]

Trick : Rationalization of denominator and put 1−x2=t2.1 - x^{2} = t^{2}.