What about using the hint?
the double integral come when i integrate by parts:
check this out for me:
u = pi - x
integrate from 0 to pi: xf(sin x) dx
= integrate from pi to 0 (pi - u)f(sin (pi - u) ) du (sub u = pi -x)
=integrate from pi to 0 uf(sin (pi - u) du - (pi)integrate from pi to 0 f(sin(pi - u) du
= (u)integrate from pi to u f(sin (pi - u)) du - integrate from pi to 0 integrate from pi to 0 f(sin (pi - u)) du - (pi) integrate from pi to 0 f(sin(pi - u) du.
the double integral is at the middle portion.
I see no reason to use integration by parts or any double integral.
Let as the hint suggests. Then . Further, when x= 0, and when u= 0 so the limits of integration are swapped. But that "-" just swaps them back again. Of couse, [itex]x= \pi- u[/ietx]. In terms of the variable u, the integral is
And, of course, so we have
But these are definite integrals- the variables are "dummy" variables- they don't exist outside the integrals. So it is perfectly valid to make the new substitution "u= x" on the right and have.
and the result follows easily.