Integral Craze 2

To prove: $\displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(a + b - x) + f(x)} dx = \dfrac{b - a}{2}$

Let $I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(a + b - x) + f(x)} dx$ and $u = a + b - x$

$\implies I = \displaystyle\int_{a}^{b} \dfrac{f(u)}{f(u) + f(a + b - u)} du$

Since $u$ is a dummy variable we can replace $u$ by $x$ to obtain:

$I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(x) + f(a + b - x)} dx$

Adding $I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(a + b - x) + f(x)} dx$ and $I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(x) + f(a + b - x)} dx$

$\implies 2I = \displaystyle\int_{a}^{b} dx \implies \boxed{I = \dfrac{b - a}{2}}$ .

Now to integrate $I = \displaystyle\int_{0}^{\dfrac{\pi}{2}} \dfrac{\sqrt{\cos(x)}}{\sqrt{\cos(x)} + \sqrt{\sin(x)}} dx$

Let $f(x) = \sqrt{\sin(x)} \implies f(\dfrac{\pi}{2} - x) = \sqrt{\cos(x)} \implies I = \displaystyle\int_{0}^{\dfrac{\pi}{2}} \dfrac{f(\dfrac{\pi}{2} - x)}{f(\dfrac{\pi}{2} - x) + f(x)} dx$

From the above result we obtain $I = \dfrac{\dfrac{\pi}{2} - 0}{2} = \dfrac{\pi}{4} \approx \boxed{0.785398}$