Just another integration problem

Let

${ I }_{ 1} =\int_{ a }^{ x } { \frac { { cos }^{ 2 }(t)+cost }{ 1+sint+cost } dt }$ ${ I }_{ 2 }=\int_{ a }^{ x } { \frac { { sin }^{ 2 }(t)+sint }{ 1+sint+cost } dt }$ Then , ${ I }_{ 1 }+{ I }_{ 2 }=\int_{ a }^{ x } { \frac { { sin }^{ 2 }(t)+sinx+{ cos }^{ 2 }t+cost }{ 1+sint+cost } dt } =\int_{ a }^{ x } { \frac { 1+sint+cost }{ 1+sint+cost } dt } =x+c$ And ${ I }_{ 2 }-{ I }_{ 1 }=\int_{ a }^{ x } { \frac { { sin }^{ 2 }(t)+sint-{ cos }^{ 2 }t-cost }{ 1+sint+cost } dx } =\int_{ a }^{ x } { \frac { sint-cost }{ 1 } dt } =-sinx-cosx+c$

Solving the two equations for ${ I }_{ 1 }$ and ${ I }_{2 }$ . We get ${ I }_{ 1 }=\frac { 1 }{ 2 } \left( x+sinx+cosx \right) +C = f\left( x\right)$ Using $f\left( 0\right)=\frac{1}{2}+\frac{\pi}{8}$ . We get $f\left( \frac{-\pi}{4} \right) = \boxed{0}$