Question

Question #d405a

Answer 1

Please see below.

Explanation:

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Answer 2

`int_0^pi (xsinx)/[1+(cosx)^2]*dx=pi^2/4`

Explanation:

`I=int_0^pi (xsinx)/[1+(cosx)^2]*dx`

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

`I=int_pi^0 [(pi-y)*sin(pi-y)(-dy)]/[1+(cos(pi-y))^2]`

=`int_pi^0 (-(pi-y)*siny*dy)/[1+(-cosy)^2]`

=`int_pi^0 (-(pi-y)*siny*dy)/[1+(cosy)^2]`

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

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

After collecting 2 integrals,

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

=`pi*int_0^pi sinx/[1+(cosx)^2]*dx`

=`pi*[-arctan(cosx)]_0^(pi)`

=`(-pi)*[arctan(cos(pi))-arctan(cos0)]`

=`(-pi)*[arctan(-1)-arctan(1)]`

=`(-pi)*[(-pi/4)-(pi/4)]`

=`(-pi)*(-pi/2)`

=`pi^2/2`

Thus, `I=int_0^pi (xsinx)/[1+(cosx)^2]*dx=pi^2/4`