Play with functions 4

Relevant wiki: Integration Tricks

Similar solution with @abdulmuttalib lokhandwala's.

$\begin{aligned} f(b) & = \int_0^\pi \frac x{b^2-\cos^2 x}dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\pi \left(\frac x{b^2-\cos^2 x} + \frac {\pi-x}{b^2-\cos^2 x} \right) dx \\ & = \frac 12 \int_0^\pi \frac \pi{b^2-\cos^2 x} dx & \small \color{#3D99F6} \text{Integral is symmetrical about }\frac \pi 2. \\ & = \int_0^\frac \pi 2 \frac \pi{b^2-\cos^2 x} dx & \small \color{#3D99F6} \text{Multiply up and down by } \sec^2 x \\ & = \int_0^\frac \pi 2 \frac {\pi \sec^2 x}{b^2\sec^2 x-1} dx & \small \color{#3D99F6} \text{Let }t = \tan x \implies dt = \sec^2 x\ dx \\ & = \int_0^\infty \frac \pi{b^2t^2+b^2-1} dt \\ & = \frac \pi{b^2-1}\int_0^\infty \frac 1{\frac {b^2}{b^2-1}t^2 + 1}dt \\ & = \frac \pi{b\sqrt{b^2-1}} \tan^{-1} \left(\frac {bt}{\sqrt{b^2-1}}\right) \Bigg|_0^\infty \\ & = \frac {\pi^2}{2b\sqrt{b^2-1}} \end{aligned}$

Therefore, $\dfrac {15}{\pi^2} f\left(\dfrac 54\right) = \dfrac {15}{\pi^2} \cdot \dfrac {\pi^2}{2\cdot \frac 54 \sqrt{\left(\frac 54\right)^2-1}} = \boxed{8}$ .

$f\left( b \right) =\int _{ 0 }^{ \pi }{ \frac { x }{ { b }^{ 2 }-{ (\cos { x) } }^{ 2 } } } \\ f\left( b \right) =\int _{ 0 }^{ \pi }{ \frac { \pi -x }{ { b }^{ 2 }-({ \cos { x) } }^{ 2 } } } \\ \\ Adding\quad both\quad we\quad get\\ \\ f\left( b \right) =\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ \frac { 1 }{ { b }^{ 2 }-{ (\cos { x) } }^{ 2 } } } \\ f\left( b \right) =\int _{ 0 }^{ \pi }{ \frac { { (\sec { x) } }^{ 2 } }{ { b }^{ 2 }{ (\sec { x) } }^{ 2 }-1 } } \\ \\ Taking\quad tanx=t\quad and\quad solving\quad we\quad get\quad \\ \\ f\left( b \right) =\frac { { \pi }^{ 2 } }{ 2b } \frac { 1 }{ \sqrt { { b }^{ 2 }-1 } }$