A calculus problem by Carwaniwer Qee

$\begin{aligned} I&=\int_0^\pi \sqrt{1-\sin x}\, dx \\&=\int_0^\pi \sqrt{\sin^2\frac x2+\cos^2\frac x2-2\sin \frac x2\cos \frac x2}\, dx \\&=\int_0^\pi \sqrt{\left(\sin\frac x2-\cos\frac x2\right)^2}\, dx \\&=\int_0^\pi\left|\sin\frac x2-\cos\frac x2\right|\, dx \\&=\int_0^{\pi /2}\left(\cos\frac x2-\sin\frac x2\right)\, dx+\int_{\pi/2}^\pi\left(\sin\frac x2-\cos\frac x2\right)\, dx \\&=\left.\left(2\sin\frac x2+2\cos\frac x2\right)\right|_0^{\pi /2}+\left.\left(-2\cos\frac x2-2\sin\frac x2\right)\right|_{\pi /2}^\pi \\&=\boxed{4\sqrt2-4} \end{aligned}$

$\begin{aligned} I = &\int_0^{\pi} \sqrt{1-\sin x}\,dx \\ \\ = &\int_0^{\pi} \frac{\sqrt{1-\sin^2 x}}{\sqrt{1+\sin x}}\, dx \\ \\ = & \int_0^{\pi}\frac{| \cos x |}{\sqrt{1+\sin x}} \, dx \\ \\ = & \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sqrt{1+\sin x}} \,dx + \int_{\frac{\pi}{2}}^{\pi} \frac{-\cos x}{\sqrt{1+\sin x}} \,dx \\ = & \int_1^2 \frac{1}{\sqrt{u}}\,du + \int_1^2 \frac{1}{\sqrt{u}}\,du \, \, \, \, \, \, \, \, \, \text{via u = 1+sin x} \\ \\ I = & 2\int_1^2\frac{1}{\sqrt{u}}\,du = 2\big( 2\sqrt{u} \big |_1^2 \big) = \boxed{4\sqrt{2}-4} \end{aligned}$

$\begin{aligned} I & = \int_0^\pi \sqrt{1-\sin x}\ dx & \small \color{#3D99F6} \text{Since integrand is symmetrical about }\frac \pi 2 \\ & = 2 \int_0^\frac \pi 2 \sqrt{1-\sin x}\ dx & \small \color{#3D99F6} \text{By half-angle tangent substitution and} \\ & = 2 \int_0^1 \sqrt{1-\frac {2t}{1+t^2}}\cdot \frac {2dt}{1+t^2} & \small \color{#3D99F6} \text{let }t = \tan \frac x2 \implies dt = \frac 12 \sec^2 \frac x2 dx \\ & = 4 \int_0^1 \frac {1-t}{(t^2+1)^\frac 32} dt \\ & = 4 \int_0^1 \left({\color{#3D99F6} \frac 1{(t^2+1)^\frac 32}} - \frac t{(t^2+1)^\frac 32}\right) dt &\small \color{#3D99F6} \text{Let }t = \tan \theta \implies dt = \sec^2 \theta \ d\theta \\ & = 4 \left[- \frac 1{\sqrt{t^2+1}}\right]_0^1 - 4 \color{#3D99F6} \int_0^\frac \pi 4 \frac {\sec^2 \theta}{\sec^3 \theta}\ d\theta \\ & = 4 \int_0^\frac \pi 4 \cos \theta\ d\theta - 4\left(1-\frac 1{\sqrt 2}\right) \\ & = 4 \sin \theta\ \bigg|_0^\frac \pi 4 - 4 + 2\sqrt 2 \\ & = 2\sqrt 2 - 4 + 2\sqrt 2 \\ & = \boxed{4\sqrt 2-4} \end{aligned}$