Calcul integral 09.11.23

Calcul

$\displaystyle I=\int_0^{\pi/2} \frac{f(\sin x) }{f(\sin x)+f(\cos x)} dx$

Solution

$u=\frac{\pi}{2}-x; du=-x$ $x=0, u=\frac{\pi}{2}$ $x=\frac{\pi}{2}, u=0$ $\displaystyle \implies I=\int_0^{\pi/2} \frac{f(\sin x )}{f(\sin x)+f(\cos x)} dx$ $\displaystyle =\int_{\pi/2}^{0} \frac{f(\sin (\pi/2-u) )}{f(\sin (\pi/2-u))+f(\cos (\pi/2-u))} (-du)$ $\displaystyle =-\int_{\pi/2}^{0} \frac{f(\cos u) }{f(\sin u)+f(\cos u)} du$ $\displaystyle =\int_0^{\pi/2} \frac{f(\cos u) }{f(\sin u)+f(\cos u)} du$ $\displaystyle =\int_0^{\pi/2} \frac{f(\cos x )}{f(\sin x)+f(\cos x)} dx$

So,

$\displaystyle I+I=\int_0^{\pi/2} \frac{f(\sin x) }{f(\sin x)+f(\cos x)} dx+\int_0^{\pi/2} \frac{f(\cos x )}{f(\sin x)+f(\cos x)} dx$ $\displaystyle 2I=\int_0^{\pi/2} \frac{f(\sin x) +f(\cos x)}{f(\sin x)+f(\cos x)} dx$ $\displaystyle 2I=\int_0^{\pi/2} dx=\frac{\pi}{2}$ $\displaystyle I=\frac{\pi}{4}$

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