Question: I'm solving some exercises about integrals but I can't solve this one, can someone explain the step-to-step please?

Let $\phi:[a,b]\rightarrow\mathbb{R}$ be a function with a second derivative, where $f(0)=0$ and $f(\pi)=3$, calculate $\int\_{0}^{\pi}[f(t)+f''(t)]sin(t)dt$

Answer: 
This is just a variation on @mickep argument. Set $p>1$, now we have
$$\frac{1}{t}\int\_0^t\frac{f(s)}{s}ds=\int\_0^1\frac{f(ty)}{ty}dy.$$
Applying Holder we get
$$\Big(\int\_0^1\frac{f(ty)}{ty}dy\Big)^p\le\int\_0^1\Big(\frac{f(ty)}{ty}\Big)^pdy$$
and integrating on both sides in $dt/t$
$$\int\_0^\infty\Big(\int\_0^1\frac{f(ty)}{ty}dy\Big)^p\frac{dt}{t}\le\int\_0^\infty\int\_0^1\Big(\frac{f(ty)}{ty}\Big)^pdy\frac{dt}{t}$$
Using Fubini on the right
$$\int\_0^\infty\Big(\int\_0^1\frac{f(ty)}{ty}dy\Big)^p\frac{dt}{t}\le\int\_0^1\int\_0^\infty\Big(\frac{f(ty)}{ty}\Big)^p\frac{dt}{t}dy$$
Finally setting $s=ty$ in the right hand side we get an integral independent from $y$ and the desired inequality.
I can prove this inequality if $f (x) /x$ were non-decreasing.

All we need to prove is $$\int\_{0}^{t} \frac {f (s)} {s} ds \leqslant f(t).$$ Take $g (x) = f (x) / x$ wherever defined, then we should prove $$\int\_{0}^{t} g (s) ds \leqslant t g(t).$$ Since $g$ is non-decreasing, we are done.