Manipulating The Integral

Make the substitution $u=\pi-x \\\frac { du }{ dx } =-1\\dx=-du\\\\u=\pi-0=\pi\\u=\pi-\pi=0$ $I=-\int _{ \pi }^{ 0 }{ \frac { (\pi-u)\sin(\pi-u) }{ 1+\cos^2(\pi-u) } du } =\int _{ 0 }^{ \pi }{ \frac { (\pi-u)\sin(\pi-u) }{ 1+\cos^2(\pi-u) } du }$ $\sin(\pi-u)=\sin u$ $cos^2(\pi-u)=\cos^2u$ $\int _{ 0 }^{ \pi }{ \frac { (\pi -u)\sin (u) }{ 1+\cos ^{ 2 } (u) } du } =\int _{ 0 }^{ \pi }{ \frac { (\pi )\sin (u) }{ 1+\cos ^{ 2 } (u) } du } -\int _{ 0 }^{ \pi }{ \frac { (u)\sin (u) }{ 1+\cos ^{ 2 } (u) } du }$ $t=-\cos u \\\frac {dt} {du} = \sin u\\du = \frac {dt} {\sin u} \\t=-\cos 0=-1\\t=-\cos \pi=1$ Therefore: $\int _{ -1 }^{ 1 }{ \frac { \pi }{ 1+t^2 } dt} =\pi(\arctan(1) - \arctan(-1))=\frac {\pi^2} {2}$ And the second integral is equivalent to: $I$ Therefore: $I=\frac {\pi^2} {2} - I\\I=\frac {\pi^2} {4}$ . $a + b + c = 1+2+4 = \boxed{7}$ .