<s>[INST] Proving $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx=\frac{π}{2}-1$ Proving $$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx= \frac{π}{2}-1$$
My attempt is: 
I assumed the $1-x=u$
$du =-dx$
$$\int_0^1 \sqrt{\frac{u}{2+u}}\,(-du)$$
 [/INST] 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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