Question

What is `int_0^pi\ x/(1+sin^2(x))`?

Answer 1

`int_0^(pi) x/(1+(sinx)^2)*dx=(pi^2sqrt2)/4`

Explanation:

`I=int_0^(pi) x/(1+(sinx)^2)*dx`

After using `x=pi-y` and `dx=-dy` transforms, `I` became

`I=int_(pi)^0 (pi-y)/(1+(sin(pi-y))^2)*(-dy)`

=`int_(pi)^0 (pi-y)/(1+(siny)^2)*(-dy)`

=`int_0^(pi) (pi-y)/(1+(siny)^2)*dy`

=`int_0^(pi) (pi-x)/(1+(sinx)^2)*dx`

After summing 2 integrals,

`2I=int_0^(pi) x/(1+(sinx)^2)*dx`+`int_0^(pi) (pi-x)/(1+(sinx)^2)*dx`

=`piint_0^(pi) (dx)/(1+(sinx)^2)`

=`piint_0^(pi) ((cscx)^2*dx)/((cscx)^2+1)`

=`piint_0^(pi) ((cscx)^2*dx)/((cotx)^2+2)`

=`piint_(pi)^0 (-(cscx)^2*dx)/((cotx)^2+2)`

After using `z=cotx` and `dz=-(cscx)^2*dx` transforms, it became

`2I=piint_(-oo)^oo (dz)/(z^2+2)`

=`(pisqrt2)/2*[arctan(z/sqrt2)]_(-oo)^oo`

=`(pi^2sqrt2)/2`

Thus, `I=(pi^2sqrt2)/4`