A calculus problem by Jose Sacramento

$\begin{aligned} I & = \int_0^\pi \sqrt{1+\sin x} \ dx & \small \color{#3D99F6}{\text{As integrand is symmetrical on }x=\frac \pi 2} \\ & = 2 \int_0^\frac \pi 2 \sqrt{1+\sin x} \ dx & \small \color{#3D99F6}{\text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = 2 \int_0^\frac \pi 2 \sqrt{1+\color{#3D99F6}{\cos x}} \ dx & \small \color{#3D99F6}{\sin \left(\frac \pi 2 - x\right) = \cos x} \\ & = 2 \int_0^\frac \pi 2 \sqrt{1+2\cos^2 \frac x2 -1} \ dx \\ & = 2 \int_0^\frac \pi 2 \sqrt2 \cos \frac x2 \ dx \\ & = 4\sqrt 2 \sin \frac x2 \ \bigg|_0^\frac \pi 2 \\ & = 4\sqrt 2 \left(\frac 1{\sqrt 2} - 0\right) \\ & = \boxed{4} \end{aligned}$

1+sinx ={ [sin(x/2)]^2+[cos(x/2)]^2 } +2sin(x/2)cos(x/2).
= (1+sinx)^1/2= sin (x/2) +cos (x/2)

On indefinitely integrating we get the integrated expression as =2sin(x/2) -2 cos (x/2)
On putting the limits in the expression above = [ (0+2) - (-2+0)] =4