Question #d405a

2 Answers
Sean
Nov 24, 2017

Please see below.

Explanation:

This is a very difficult integral to calculate. It is not possible to enter the solution in this space. Please click on the link below and enter your function to see the steps and final answer.

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Cem Sentin
Nov 24, 2017

#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#

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