$$\int_0^{\frac{\pi}{2}}x\left(\frac{\sin x}{1+\cos^2 x}\right)dx$$ integralinde $x$ yerine $\dfrac{\pi}{2}-x$ alırsak bu integral $$\int_0^{\frac{\pi}{2}}\left(\frac{(\pi/2-x)\cos x}{1+\sin^2 x}\right)dx$$ şekline dönüşür. Buna göre $$\int_0^{\frac{\pi}{2}}x\left(\frac{\sin x}{1+\cos^2 x}\right)dx+\int_0^{\frac{\pi}{2}}x\left(\frac{\cos x}{1+\sin^2 x}\right)dx=\int_0^{\frac{\pi}{2}}\left(\frac{(\pi/2-x)\cos x}{1+\sin^2 x}\right)dx+\int_0^{\frac{\pi}{2}}x\left(\frac{\cos x}{1+\sin^2 x}\right)dx$$ $$=(\pi/2)\int_0^{\frac{\pi}{2}}\left(\frac{\cos x}{1+\sin^2 x}\right)dx$$ elde edilir.
$t=\sin x$ dönüşümü ile $$(\pi/2)\int_0^{1}\dfrac{dt}{1+t^2}=(\pi/2)\arctan t\Big|_0^1=\dfrac{\pi^2}{8}$$ bulunur.