توجه کنید که:
$$ \int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
$$,ln^2(cotx)=ln^2( \frac{1}{tanx})=(ln1-ln(tanx))^2=(0-ln(tanx))^2=ln^2(tanx)$$
$$ \Rightarrow I:=\int_0^ \frac{\pi}{2} \frac{tanx}{4ln^2(tanx)+\pi^2}dx= \int_0^ \frac{\pi}{2} \frac{tan( \frac{\pi}{2} -x)}{4ln^2(tan( \frac{\pi}{2} -x))+\pi^2}dx$$
$$=\int_0^ \frac{\pi}{2} \frac{cotx}{4ln^2(cotx)+\pi^2}dx=\int_0^ \frac{\pi}{2} \frac{cotx}{4ln^2(tanx)+\pi^2}dx$$
$$ \Rightarrow 2I=\int_0^ \frac{\pi}{2}\frac{tanx+cotx}{4ln^2(tanx)+\pi^2}dx=\int_0^ \frac{\pi}{2} \frac{tanx}{tanx}.\frac{tanx+cotx}{4ln^2(tanx)+\pi^2}dx$$
$$=\int_0^ \frac{\pi}{2} \frac{1+tan^2x}{tanx(4ln^2(tanx)+\pi^2)}dx$$
حالا تغییر متغیر $u:= \frac{2}{\pi}ln(tanx)$ را به کار بگیرید:
$$ \Rightarrow du= \frac{2}{\pi}.\frac{1+tan^2x}{tanx}dx$$
$$,2I= \frac{\pi}{2}\int_{- \infty }^ \infty \frac{1}{(4 \frac{\pi^2u^2}{4}+\pi^2)}du=2 \times \frac{\pi}{2}. \frac{1}{\pi^2}\int_0^ \infty \frac{1}{1+u^2}du$$
$$= \frac{1}{\pi}tan^{-1}x|_{x=0}^{x= \infty }=\frac{1}{\pi}( \frac{\pi}{2}-0)= \frac{1}{2}$$
$$ \Rightarrow I=\frac{1}{4}$$
$\Box$
