int_0^1 sqrt{frac{1-x}{1+x}} , dx is equal to | vedclass

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(A) Let $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$. Rationalizing the integrand, we get: $I = \int_0^1 \sqrt{\frac{1-x}{1+x} 	imes \frac{1-x}{1-x}}

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

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(A) Let $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$. Rationalizing the integrand, we get: $I = \int_0^1 \sqrt{\frac{1-x}{1+x} 	imes \frac{1-x}{1-x}} \, dx = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \, dx$. Splitting the integral: $I = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$. Evaluating the first part: $\int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx = [\sin^{-1} x]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. Evaluating the second part: Let $u = 1-x^2$, then $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} du$. $\int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int_1^0 u^{-1/2} \, du = \frac{1}{2} \int_0^1 u^{-1/2} \, du = \frac{1}{2} [2\sqrt{u}]_0^1 = [\sqrt{u}]_0^1 = 1 - 0 = 1$. Thus, $I = \frac{\pi}{2} - 1$.

$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$ is equal to

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