int_0^1 sqrt{frac{1-x}{1+x}} ,dx equals | 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 \frac{\sqrt{1-x}}{\sqrt{1+x}} \cdot \frac{\sqrt{

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$\left(\frac{\pi}{2} - 1\right)$

Vedclass · Answer

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

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

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