M2 Integration

13（C)

Let I = the integration

put y = 2 pi -x , so x = 2 pi - y, dy/dx = -1

when x=5/4 pi, y = 2 pi - 5/4 pi , ie y = 3/4 pi
when x=3/4 pi, y = 2 pi - 3/4 pi , ie y =5/4 pi

Also f(x) = f( 2pi - y) = - f(y) ( by definition f(2pi -x ) = -f(x) )

but sin (x) = sin ( 2 pi - y) = -sin y ( since sin ( 2 pi - x ) = -sin x )

thus
ln( 1 + e^sin x )
= ln(1+e^(-sin y))
= ln ( ( e^sin y + 1 ) / (e^sin y ) )
= ln ( e^ sin y + 1 ) - ln ( e^sin y)
= ln ( 1 + e^sin y ) - sin y

therefore the integration becomes

{ 3/4pi
I = { -f(y)( ln(1+e^sin y) - sin y) ( -dy)
{ 5/4pi

{ 3/4pi
= { f(y) ln( e^sin y) dy - f(y) sin y dy
{ 5/4pi

{ 5/4 pi
={ -f(y) ln( 1+e^sin y) dy + f(y) sin y dy
{ 3/4 pi
( while exchange index will time -1 to the function)

{ 5/4 pi
={ -f(x) ln ( 1+e^sin x ) dx + f(x) sin x dx ( change notation y to x )
{3/4 pi

{5/4 pi
= -I + { f(x) sin x dx
{ 3/4 pi

{ 5/4 pi
= 1/2 { f(x) sin x dx
{ 3/4 pi

( Q.E.D.)

*****************************************************
13(d)

put f(x) = sin x / ( 1 + 3 (cos x)^2 )
put x = 2 pi - x
f(2pi - x)
= sin (2 pi - x ) / ( 1 + 3 ( cos (2 pi - x )^2 )
= - sin x / ( 1 + 3 ( cos x ) ^ 2 )
= - f(x )

Hence , by 13(c)

the integration
{ 5/4 pi
={ f(x) * sin x dx
{ 3/4 pi

{ 5/4 pi
= { sin x / ( 1 + 3 (cos x)^2) * sin x dx
{ 3/4 pi

{ 5/4 pi
= { ( sin x )^2 / ( 1 + 3(cos x)^2 ) dx
{ 3/4 pi

{ 5/4 pi
= { ( 1 - (cos x)^2 ) / ( 1 + 3 (cos x)^2 ) dx
{ 3/4 pi

{ 5/4 pi
={ 1 / ( 1+ 3 (cos x) ^2 dx + ( cos x ) ^2 / ( 1 + 3(cos x)^2) dx
{ 3/4 pi

{ 5/4 pi
={ 1/(1+3 (cos x)^2 dx + 1 / ( (sec x)^2 + 2 ) dx
{ 3/4 pi

so, the former part using result of 13 b(ii) and the latter part using result of a(i) with putting x with index, result follows.