Evaluate ∫0^π(xsin x)/(1+cos ^2x) dx.

Use the property $\int_0^{\pi}g(x)\,dx=\int_0^{\pi}g(\pi-x)\,dx$. Since $\sin(\pi-x)=\sin x$ and $\cos^{2}(\pi-x)=\cos^{2}x$, we get $I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^{2}x}dx$. Adding the two forms gives $2I=\pi\int_0^{\pi}\frac{\sin x}{1+\cos^{2}x}dx$. Put $u=\cos x$ so that the integral becomes $\int_{-1}^{1}\frac{du}{1+u^{2}}=\frac\pi2$. Hence $2I=\frac{\pi^{2}}{2}$ and $I=\frac{\pi^{2}}{4}\approx2.4674$. A numerical quadrature gives $2.46740$, matching. Rounded to two decimals the answer is $2.47$. The integrand has no elementary antiderivative in the original form, so the reflection trick is essential; trying integration by parts leads nowhere because the factor $x$ cannot be removed. The reflected form removes $x$ completely and reduces the problem to a standard arctangent integral, which is the usual pattern for integrals over $[0,\pi]$ with an even function of $\cos x$.