Integral with Inverse Trigo

Since the range of $\arcsin(x)$ is $\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$ and of $\arccos(x)$ is $\left[ 0, \pi \right]$ .

Hence ${ \sum _{ i=1 }^{ 6 }{ \left( \arcsin ( { x }_{ i } ) +\arccos ( { y }_{ i } ) \right) } =9\pi}$ is satisfied only when $\begin{cases} \arcsin(x_1)=\arcsin(x_2)=\arcsin(x_3)=\arcsin(x_4)=\arcsin(x_5)=\arcsin(x_6)=\dfrac{\pi}{2} \\ \arccos(y_1)=\arccos(y_2)=\arccos(y_3)=\arccos(y_4)=\arccos(y_5)=\arccos(y_6)=\pi \end{cases}$

And to achieve the above situation, we have to conclude that $\begin{cases} x_1=x_2=x_3=x_4=x_5=x_6=1 \\ y_1=y_2=y_3=y_4=y_5=y_6=-1\end{cases}$

Therefore we can evaluate $\displaystyle \sum _{ i=1 }^{ 6 }{ { x }_{ i } }=6$ and $\displaystyle \sum _{ i=1 }^{ 6 }{ { y }_{ i } }=-6$ . Therefore the given integral becomes $\displaystyle \int \limits_{ 6 }^{ -6 }{ x\ln { (1+{ x }^{ 2 }) } } \left( \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } \right) \ \mathrm{d}x$

And we know that $\left[ { x\ln { (1+{ x }^{ 2 }) } } \left( \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } \right) \right]$ is an odd function.

Hence using the property of the definite integrals that $\displaystyle \int_{-a}^{a} f(x) dx=0$ , if $f(x)$ is an odd function, we can say that $\displaystyle \int \limits_{ \displaystyle \sum _{ i=1 }^{ 6 }{ { x }_{ i } } }^{ \displaystyle \sum _{ i=1 }^{ 6 }{ { y }_{ i } } }{ x\ln { (1+{ x }^{ 2 }) } } \left( \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } \right) \ \mathrm{d}x =0$

enjoy!