Q92 / 160 · TS EAMCET 2015 · Mathematics

$\int_0^1 \sqrt{\frac{1-x}{1+x}} d x$ is equal to

  • Correct answer
    A
    $\frac{\pi}{2}-1$
  • B
    $\frac{\pi}{2}+1$
  • C
    $\pi-1$
  • D
    $\frac{3 \pi}{2}$

Solution

Let $$ \begin{aligned} I & =\int_0^1 \sqrt{\frac{1-x}{1+x}} d x \\ & =\int_0^1 \sqrt{\frac{1-x}{1+x}} \times \sqrt{\frac{1-x}{1-x}} d x \\ & =\int_0^1 \frac{1-x}{\sqrt{1-x^2}} d x \\ & =\int_0^1 \frac{1}{\sqrt{1-x^2}} d x-\int_0^1 \frac{x}{\sqrt{1-x^2}} d x \end{aligned} $$ $$ \begin{aligned} & =\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1 \\ & =\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+(\sqrt{1-1}-\sqrt{1-0}) \\ & =\frac{\pi}{2}-1 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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