We splitsen eerst de bepaalde integraal in de volgende twee integralen :
INT = \( \int_0^\frac{\pi}{2} \frac {\cos^4x} {\cos^4x+ \sin^4x} \;dx + \int_\frac{\pi}{2}^{\pi} \frac {\cos^4x} {\cos^4x+ \sin^4x} \;dx\\ \)
In de tweede integraal passen we de volgende substitutie toe :
\(y = x - \frac {\pi} {2} \quad (\scriptsize nieuwe\;ondergrens\;0,\; nieuwe\;bovengrens\;\frac {\pi}{2}, \; dy = dx)\\ \cos^4(y+\frac{\pi}{2})=\sin^4(\frac{\pi}{2}-y-\frac{\pi}{2}) =\sin^4(-y)=\sin^4 y\\ \sin^4(y+\frac{\pi}{2})=\cos^4(\frac{\pi}{2}-y-\frac{\pi}{2}) =\cos^4(-y)=\cos^4 y\\ \)
De tweede integraal wordt dan :
\( \int_0^\frac{\pi}{2} \frac {\sin^4 y} {\sin^4y+ \cos^4 y} \;dy \overset{!!}{=}\int_0^\frac{\pi}{2} \frac {\sin^4 x} {\sin^4 x+ \cos^4 x} \;dx \)  zodat
INT = \( \int_0^\frac{\pi}{2}\frac {\cos^4x} {\cos^4x+ \sin^4x} \;dx + \int_0^\frac{\pi}{2} \frac {\sin^4x} {\cos^4x+ \sin^4x} \;dx \\ = \int_0^\frac{\pi}{2}\frac {\cos^4 x + sin^4 x} {\cos^4x+ \sin^4x} \;dx = \int_0^\frac{\pi}{2} 1\;dx = \frac {\pi }{2} \)