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प्रश्न
उत्तर
\[Let\ I = \int_0^1 \sqrt{\frac{1 - x}{1 + x}} d\ x . Then, \]
\[I = \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \times \frac{\sqrt{1 - x}}{\sqrt{1 - x}} d x\]
\[ \Rightarrow I = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} dx\]
\[ \Rightarrow I = \int_0^1 \frac{1}{\sqrt{1 - x^2}} dx - \int_0^1 \frac{x}{\sqrt{1 - x^2}} dx\]
\[ \Rightarrow I = \left[ \sin^{- 1} x \right]_0^1 + \frac{1}{2} \int_0^1 \frac{- 2x}{\sqrt{1 - x^2}} dx\]
\[ \Rightarrow I = \left[ \sin^{- 1} x \right]_0^1 + \frac{1}{2} \left[ 2\sqrt{1 - x^2} \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{2} - 0 + 0 - 1\]
\[ \Rightarrow I = \frac{\pi}{2} - 1\]
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