Question

$$\int_0^{\frac{\pi}{2}}x\left(\frac{\sin x}{1+\cos^2 x}+\frac{\cos x}{1+\sin^2 x}\right)dx=?$$

Answer 1

$$\int_0^{\frac{\pi}{2}}x\left(\frac{\sin x}{1+\cos^2 x}\right)dx$$  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.