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TS EAMCET · Maths · Definite Integration

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

  1. A \(\frac{\pi}{2}-1\)
  2. B \(\frac{\pi}{2}+1\)
  3. C \(\pi-1\)
  4. D \(\frac{3 \pi}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\pi}{2}-1\)

Step-by-step Solution

Detailed explanation

Let \[ \begin{aligned} I & =\int_0^1 \sqrt{\frac{1-x}{1+x}} d x \\ & =\int_0^1 \sqrt{\frac{1-x}{1+x}} \times \sqrt{\frac{1-x}{1-x}} d x \\ & =\int_0^1 \frac{1-x}{\sqrt{1-x^2}} d x \\ & =\int_0^1 \frac{1}{\sqrt{1-x^2}} d x-\int_0^1 \frac{x}{\sqrt{1-x^2}} d x \end{aligned} \]…