Thread: Proving Question - Stucked

Hi, i am stucked with this question after stuck at double integral part. Any help would be appreciated. Thank you very much

What about using the hint?

i used the hint but it doesnt seem to get me anywhere, i show u which part i stucked at lol
after i scanned.......

wonder where the double integral came in here???

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 $\displaystyle u= \pi- x$ as the hint suggests. Then $\displaystyle du= -dx$. Further, when x= 0, $\displaystyle u= \pi$ and when $\displaystyle x= \pi$ 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
$\displaystyle \int_0^\pi (\pi- u)f(sin(\pi- u) du= \pi\int_0^\pi f(sin(\pi- u))du- \int_0^\pi u f(sin(\pi- u))du$

And, of course, $\displaystyle sin(\pi- u)= sin(u)$ so we have
$\displaystyle \int_0^\pi xf(sin(x))dx= \pi\int_0^\pi f(sin(u))du- \int_0^\pi uf(sin(u))du$

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.
$\displaystyle \int_0^\pi xf(sin(x))dx= \pi\int_0^\pi f(sin(x))dx- \int_0^\pi xf(sin(x))dx$
and the result follows easily.

Thanks alot haha, I didnt see that part