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प्रश्न
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
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उत्तर
\[\int_0^1 \sqrt{\frac{1 - x}{1 + x}} d x\]
\[ = \int_0^1 \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}} d x\]
\[ = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} d x\]
\[ = \int_0^1 \frac{1}{\sqrt{1 - x^2}}dx - \int_0^1 \frac{x}{\sqrt{1 - x^2}}dx\]
\[ = \left[ \sin^{- 1} x \right]_0^1 + \left[ \sqrt{1 - x^2} \right]_0^1 \]
\[ = \frac{\pi}{2} - 1\]
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