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Proving $$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx= \frac{π}{2}-1$$ My attempt is:

I assumed the $1-x=u$ $du =-dx$ $$\int_0^1 \sqrt{\frac{u}{2+u}}\,(-du)$$

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let $x=\cos u$ so that $dx=-\sin u \, du$ \begin{align} & =\int_{\pi /2}^{0}\sqrt{\frac{1-\cos u}{1+\cos u}}(-\sin u) \, du \\ & =\int_{0}^{\pi /2}\tan (u/2)\sin u \, du \\ & =\int_{0}^{\pi/2}\frac{\sin u}{1+\cos u}\sin u \, du=\int_0^{\pi/2} \frac{1-\cos^2 u}{1+\cos u} \, du \\ & =\int_{0}^{\pi/2}(1-\cos u) \, du=\pi/2-1-(0-0)=\frac{\pi}{2}-1 \end{align}

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Multiply both numerator and denominator by $\sqrt{1-x}$, then $\sqrt{\frac{1-x}{1+x}}$ becomes $\frac{1-x}{\sqrt{1-x^2}}$. The integral of first term is $\arcsin x$, the second term can be integrated by substitution.

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$$ \begin{align} \int_0^1\sqrt{\frac{1-x}{1+x}}\,\mathrm{d}x &=2\int_{1/2}^1\sqrt{\frac{1-x}{x}}\,\mathrm{d}x\tag{1}\\ &=4\int_{1/\sqrt2}^1\sqrt{1-x^2}\,\mathrm{d}x\tag{2}\\ &=4\int_0^{\pi/4}\sin^2(x)\,\mathrm{d}x\tag{3}\\ &=2\int_0^{\pi/4}1-\cos(2x)\,\mathrm{d}x\tag{4}\\ &=\left[2x-\sin(2x)\vphantom{\int}\right]_0^{\pi/4}\tag{5}\\[6pt] &=\frac\pi2-1\tag{6} \end{align} $$ Explanation:
$(1)$: $x\mapsto2x-1$
$(2)$: $x\mapsto x^2$
$(3)$: $x\mapsto\cos(x)$
$(4)$: $\cos(2x)=1-2\sin^2(x)$
$(5)$: integrate
$(6)$: evaluate

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Let $u = \sqrt{\dfrac{1-x}{1+x}} \to u^2 = \dfrac{1-x}{1+x} \to x = \dfrac{1-u^2}{1+u^2} \to dx = \dfrac{-4u}{(1+u^2)^2}du \to I = \displaystyle \int_{0}^1\dfrac{4u^2}{(1+u^2)^2}du$. Next let $u = \tan \theta \to du = \sec^2 \theta d\theta \to I = \displaystyle \int_{0}^{\pi/4} \dfrac{4\tan^2 \theta}{\sec^4 \theta}\cdot \sec^2 \theta\cdot d\theta = \displaystyle \int_{0}^{\pi/4} 4\sin^2 \theta d\theta = \displaystyle \int_{0}^{\pi/4} 2-2\cos (2\theta) d\theta = \pi/2-1$

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