The value of the integral int_0^1 sqrt{frac{1 | vedclass

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(A) To evaluate the integral $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the integrand:Multiply the numerator and denominator by $\sqrt

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

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(A) To evaluate the integral $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the integrand:Multiply the numerator and denominator by $\sqrt{1-x}$:$I = \int_0^1 \frac{1-x}{\sqrt{(1+x)(1-x)}} \, dx = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \, dx$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 integral,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$.The first part is $\left[ \sin^{-1}(x) \right]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.The second part is $\int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx = \left[ -\sqrt{1-x^2} \right]_0^1 = -(\sqrt{0} - \sqrt{1}) = 1$.Combining these,$I = \frac{\pi}{2} - 1$.

The value of the integral $\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$ is

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