Integration and continuity by Dimitris

• if g fraction ,which is defined $R$ -{0,π} . g(x)= $\frac{f(x)}{sinx}$ , =>f(x)=g(X) sinx , because $\displaystyle \lim_{x\to0}\dfrac{f(x)}{\sin x} = 1$ => $\displaystyle \lim_{x\to0}\ g(x) = 1$

• $\displaystyle \lim_{x\to0}\ f(x) =$ $\displaystyle \lim_{x\to0}\ g(x) sinx = g(0) sin0$ =0,because f fraction is continuous $\displaystyle \lim_{x\to0}\ f(x) = f(0)$

• f fraction is differentiable $\displaystyle \lim_{x\to0}\dfrac{f(x)-f(0)}{\ x-0} = f΄(0)$ = $\displaystyle \lim_{x\to0}\dfrac{g(x) sinx}{\ x} =$ $\displaystyle \lim_{x\to0}\dfrac{(g(x) sinx)΄}{( x)΄} =$ $\displaystyle \lim_{x\to0}\ g΄(x) sinx +g(x) cosx =$ g΄(0) sin0 +g(0) cos0=1

(because it arises limit $\frac{0}{0}$ then we can apply differentiation rule of de l'hopital)

$\displaystyle \int_0^\pi (f(x) + f'(x)) \sin x \, dx =\pi$ -=> $\displaystyle \int_0^\pi f(x) sinx + f'(x) \sin x \, dx =\pi$ => $\displaystyle \int_0^\pi (f(x) sinx) , dx + \displaystyle \int_0^\pi ( f'(x) \sin x )\, dx =\pi$ => $\displaystyle \int_0^\pi (f(x) (-cosx ) )\, dx + \displaystyle \int_0^\pi ( f'(x))' sin x \, dx =\pi$ =>

\ \left|f(x) (-cosx)\right|_{\color{}{0}}^{\color{}{\pi}} + \displaystyle \int_0^\pi ( f'(x)) (cos x) \, dx +[\ \left|f'(x) (sinx)\right|_{\color{}{0}}^{\color{}{\pi}}- \displaystyle \int_0^\pi ( f'(x)) cos x \, dx =\pi => $(-cos π) f(π)-(-cos 0) f(0)+f'(π) sin π - f(0) sin 0=π=>f(π)=π$

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