IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - INTEGRATE THE FUNCTION- f(x)= xf(cosx) FROM 0 to 2 pie ?

rohith291991

ok i think ive got it.... EXCELLENT sum btw...

NOTE : replace the question marks by integral signs....
given

   f(x)=xf(cosx)-----1

now f(-x)=-xf(cosx)
      or f(-x)=-f(x)
hence it is an odd function...

now
f(2pi-x)=(2pi-x)f(cos(2pi-x) =(2pi-x)f(-cosx)=(2pi-x)f(cosx)
(becoz its an odd function)

but f(cosx)=f(x)/x

hence we get f(2pi-x)=(2pi-x)f(x)/x

adding f(x) to both sides....

we get f(2pi-x)+f(x)=2pif(x)/x ---------2

     now integrating both sides of 2 between 0 and 2pi....

      ₀?^2pi f(2pi-x)dx + ₀?^2pi f(x)dx = 0?2pi 2pif(x)/xdx

but from properties of definite integrals we know that
   _a?^b f(a+b-x)dx= _a?^b f(x)dx

hence we get ₀?^2pi f(2pi-x)dx=₀?^2pi f(x)dx

substituting in 2 we get 2 ₀?^2pi f(x)dx=₀?^2pi 2pif(x)/xdx

or ₀?^2pi f(x)dx=₀?^2pi pif(x)/xdx ----3

now put x=x+pi (change of variable)

hence limits change....

we get ₀?^2pi f(x)dx=_-pi?^pi pif(pi+x)/(pi+x)dx

but f(pi+x)=(pi+x)f(x)/x

hence ₀?^2pi f(x)dx=_-pi?^pi pif(x)/xdx

but this is an even function.... hence _-pi?^pi pif(x)/xdx
= 2₀?^pi pif(x)/xdx now put x=x+pi/2

we get ₀?^2pi f(x)dx=2pi_-pi/2?^pi/2 f(x+pi/2)/(x+pi/2)dx

but f(x+pi/2)=(x+pi/2)f(-sinx) = -(x+pi/2)f(sinx)

hence ₀?^2pi f(x)dx=-2pi_-pi/2?^pi/2 f(sinx)dx but f(sinx) is an odd function hence _-pi/2?^pi/2 f(sinx)dx =0 hence ₀?^2pi f(x)dx=0

hence the answer is zero....

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