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Question: \(\int_{}^{}{\frac{\sin^{- 1}\sqrt{x} - \cos^{- 1}\sqrt{x}}{\sin^{- 1}\sqrt{x} + \cos^{- 1}\sqrt{x}}...

∫sin⁡−1x−cos⁡−1xsin⁡−1x+cos⁡−1xdx\int_{}^{}{\frac{\sin^{- 1}\sqrt{x} - \cos^{- 1}\sqrt{x}}{\sin^{- 1}\sqrt{x} + \cos^{- 1}\sqrt{x}}dx} is equal to

A

2π[(2x−1)sin⁡−1x+x(1−x)]−x+c\frac{2}{\pi}\lbrack(2x - 1)\sin^{- 1}\sqrt{x} + \sqrt{x(1 - x)}\rbrack - x + c

B

2π[(2x−1)sin⁡−1x−x(1−x)]+x+c\frac{2}{\pi}\lbrack(2x - 1)\sin^{- 1}\sqrt{x} - \sqrt{x(1 - x)}\rbrack + x + c

C

π2[(2x−1)sin⁡−1x+x(1−x)]−x+c\frac{\pi}{2}\left\lbrack (2x - 1)\sin^{- 1}\sqrt{x} + \sqrt{x(1 - x)} \right\rbrack - x + c

D

None of these

Answer

2π[(2x−1)sin⁡−1x+x(1−x)]−x+c\frac{2}{\pi}\lbrack(2x - 1)\sin^{- 1}\sqrt{x} + \sqrt{x(1 - x)}\rbrack - x + c

Explanation

Solution

∫sin⁡−1x−cos⁡−1xsin⁡−1x+cos⁡−1xdx=2π[∫sin⁡−1xdx−∫cos⁡−1xdx]\int \frac { \sin ^ { - 1 } \sqrt { x } - \cos ^ { - 1 } \sqrt { x } } { \sin ^ { - 1 } \sqrt { x } + \cos ^ { - 1 } \sqrt { x } } d x = \frac { 2 } { \pi } \left[ \int \sin ^ { - 1 } \sqrt { x } d x - \int \cos ^ { - 1 } \sqrt { x } d x \right] [∵sin⁡−1x+cos⁡−1x=π2]\left\lbrack \because\sin^{- 1}\sqrt{x} + \cos^{- 1}\sqrt{x} = \frac{\pi}{2} \right\rbrack

Now we solve first and second expressions separately. For first expression, ∫sin⁡−1xdx\int_{}^{}{\sin^{- 1}\sqrt{x}dx}

Put x=sin⁡2θx = \sin^{2}\theta ⇒cos⁡2θ=1−2x\Rightarrow \cos 2\theta = 1 - 2x ⇒dx=sin⁡2θdθ\Rightarrow dx = \sin 2\theta d\theta

∫θ.sin⁡2θdθ=−θcos⁡2θ2+12∫cos⁡2θdθ\int_{}^{}{\theta.\sin 2\theta d\theta = \frac{- \theta\cos 2\theta}{2} + \frac{1}{2}\int_{}^{}{\cos 2\theta}d\theta} =−θcos⁡2θ2+sin⁡2θ4+c1= \frac{- \theta\cos 2\theta}{2} + \frac{\sin 2\theta}{4} + c_{1} =(2x−1)2sin⁡−1x+12x(1−x)+c1= \frac{(2x - 1)}{2}\sin^{- 1}\sqrt{x} + \frac{1}{2}\sqrt{x(1 - x)} + c_{1}For second expression, ∫cos⁡−1xdx\int_{}^{}{\cos^{- 1}\sqrt{x}}dx

Put x=cos⁡2θx = \cos^{2}\theta ⇒dx=−2cos⁡θsin⁡θdθ=−sin⁡2θdθ\Rightarrow d x = - 2 \cos \theta \sin \theta d \theta = - \sin 2 \theta d \theta ∫cos⁡−1xdx=−∫θ.sin⁡2θdθ=θcos⁡2θ2−sin⁡2θ4+c2\int_{}^{}\cos^{- 1}\sqrt{x}dx = - \int_{}^{}{\theta.\sin 2\theta}d\theta = \frac{\theta\cos 2\theta}{2} - \frac{\sin 2\theta}{4} + c_{2}

Therefore,I=2π[(2x−1)2{sin⁡−1x−cos⁡−1x}+x(1−x)]+cI=2π(2x−1)sin⁡−1x−(2x−1)2.π2 +x(1−x)]+c⇒I=2π[(2x−1)sin⁡−1x+x(1−x)]−x+cI = \frac{2}{\pi}\left\lbrack \frac{(2x - 1)}{2}\{\sin^{- 1}\sqrt{x} - \cos^{- 1}\sqrt{x}\} + \sqrt{x(1 - x)} \right\rbrack + cI = \frac{2}{\pi}(2x - 1)\sin^{- 1}\sqrt{x} - \frac{(2x - 1)}{2}.\frac{\pi}{2}\left. \ + \sqrt{x(1 - x)} \right\rbrack + c \Rightarrow I = \frac{2}{\pi}\left\lbrack (2x - 1)\sin^{- 1}\sqrt{x} + \sqrt{x(1 - x)} \right\rbrack - x + c