int_0^1 sqrt{frac{1-x}{1+x}} , dx = | vedclass

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(D) To evaluate the integral $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the integrand: $I = \int_0^1 \frac{\sqrt{1-x}}{\sqrt{1+x}} \cd

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$\frac{\pi}{2}-1$

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(D) To evaluate the integral $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the integrand: $I = \int_0^1 \frac{\sqrt{1-x}}{\sqrt{1+x}} \cdot \frac{\sqrt{1-x}}{\sqrt{1-x}} \, dx = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \, dx$ Split the integral into two parts: $I = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$ For the second part,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$: $I = [\sin^{-1}(x)]_0^1 + \frac{1}{2} \int_1^0 u^{-1/2} \, du$ $I = [\sin^{-1}(x)]_0^1 + [\sqrt{1-x^2}]_0^1$ $I = (\sin^{-1}(1) - \sin^{-1}(0)) + (\sqrt{1-1^2} - \sqrt{1-0^2})$ $I = (\frac{\pi}{2} - 0) + (0 - 1) = \frac{\pi}{2} - 1$

$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx =$

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