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Question: If I = \(\int_{0}^{1}x\sqrt{\frac{1 - x}{1 + x}}\)dx, the I equals-...

If I = ∫01x1−x1+x\int_{0}^{1}x\sqrt{\frac{1 - x}{1 + x}}dx, the I equals-

A

1 + π4\frac{\pi}{4}

B

1 –π4\frac{\pi}{4}

C

p

D

p –2\sqrt{2}

Answer

1 –π4\frac{\pi}{4}

Explanation

Solution

We can write

I = ∫01x(1−x)1−x2\int_{0}^{1}\frac{x(1 - x)}{\sqrt{1 - x^{2}}}dx

= ∫01(x1−x2−11−x2+1−x21−x2)\int_{0}^{1}\left( \frac{x}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}} + \frac{1 - x^{2}}{\sqrt{1 - x^{2}}} \right)dx

 (−1−x2−sin⁡−1x)]01\left. \ \left( - \sqrt{1 - x^{2}} - \sin^{- 1}x \right) \right\rbrack_{0}^{1}

 (x21−x2+12sin⁡−1x)]01\left. \ \left( \frac{x}{2}\sqrt{1 - x^{2}} + \frac{1}{2}\sin^{- 1}x \right) \right\rbrack_{0}^{1}

– π2\frac{\pi}{2}+ 1 + 12\frac { 1 } { 2 } (π2)\left( \frac{\pi}{2} \right)= 1 – π4\frac{\pi}{4}