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

puneet.jas

INTEGRATE THE FUNCTION- f(x)= xf(cosx) FROM 0 to 2 pie ?

man111

f(x)=_{[0 ]}

^{[2pi ]} xf(cosx)dx

put x=2pi-x in yhis integration.

f(2pi-x)=_{0 ]}^{[2pi ]} (2pi-x)f(cos(2pi-x))dx=2pi ₀^{[2pi ]}f(cosx)dx-_{0 ]}^{[2pi ]} xf(cosx)dx

f(x)=f(2pi-x)=2pi ₀^{[2pi ]}f(cosx)dx-f(x)

2f(x)=2pi ₀^{[2pi ]}f(cosx)dx=

f(x)=₀^{[2pi ]}f(cosx)dx=2₀^{[pi ]}f(cosx)dx=

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....

rohith291991

just replace the question marks wth integral symbols...

rohith291991

yes i think i have got it plz check above post...

puneet

hii

well i went through the solution .. looks great .. i think it is all correct ..

i must congratulate u for the soln .. good work

cheers

rohith291991

thank u!!!

feynmann

Giving u a simpler soln .

Let g'( cosx ) = f ( cos x) ( It can always be written )

so we have ,

_{[0 ]}

^{[ 2pi]} f ( cos x ) dx = g ( cos ( 2pi ) ) - g ( cos ( 0 )) = 0 ...................(1 )

Now let I = given integral

change x to 2 pi - x. so the integral becomes

I =

_{[ 0]}

^{[2pi ]} 2pi f ( cos x ) dx - I

( since f( cos ( 2pi -x ) )= f ( cos x ) )

So we get I = pi

f ( cos x ) dx = 0 ( by 1 )

rohith291991

@feynmann see integral f(x) is asked not integral f(cosx).....hope u understood the flaw...

feynmann

Hey , we have to integrate x f( cos x ) right ?

I have done the integration.

rohith291991

ok yes ur right there my mistake.....

but something seems to be wrong in this step..._{[0 ]}

^{[ 2pi]} f ( cos x ) dx = g ( cos ( 2pi ) ) - g ( cos ( 0 )) = 0 is this valid for any f ?? just plot the function e^cos(x) and see http://www.pa.uky.edu/~phy211/graph_applets/plot_graph.html

feynmann

see trigonometric fn s. are clsed under the opn diff & integration .

so I can always write g'( cos x ) = f( cos x )

or , equivalently indefinite integral of f( cos x ) dx = fn of cosx ( see trig fns are inter convertible )

The point will become clear when u replace cos x by Re ( e^ ix ) whose differentiation & integration yields the same fn .

The part that u have mentioned is nothing but an application of fundamental theorem .

Hope , u got it !!!!!!!!

rohith291991

but on plotting and checking we dont get zero area thats what im saying....i just arbitrarily chose e^cosx plot it and see for urself the area is not zero between zero and 2pi....plz explain this anomaly.... but wat im asking is how does the area always equal zero......

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