Find $\int_0^\pi f_{(x)}dx$ when $f_{(x)}=\int_0^x\frac{\sin t}{t}dt$ using $f_{(\pi)}=\beta$ There's a hint included in this question to use integration by parts.
Find $\displaystyle\int_0^\pi f_{(x)}dx$ when $f_{(x)}=\displaystyle\int_0^x\frac{\sin t}{t}dt$ using $f_{(\pi)}=\beta$
Here are my attempts:
$$\displaystyle\int_0^\pi f_{(x)}dx=\displaystyle\int_0^\pi 1(\displaystyle\int_0^x\frac{\sin t}{t})dx$$
$$u' = 1, v=\displaystyle\int_0^x\frac{\sin t}{t}=f_{(x)}$$
$$u = x, v' =f'_{(x)}$$
($v'=f_{(x)}$ because  $(F_{(x)}-F_{(0)})'=f_{(x)}-f_{(0)}=f_{(x)}$)
$$\displaystyle\int_0^\pi f_{(x)}dx=uv-\displaystyle\int_0^\pi uv'dx=\pi f_{(\pi)}-\displaystyle\int_0^\pi xf'_{(x)}dx$$
Using integration by parts
$$\pi f_{(\pi)}-(\pi f_{(\pi)}-\displaystyle\int_0^\pi f_{(x)}dx)$$
Which leads me to a trivial $$\displaystyle\int_0^\pi f_{(x)}dx=\displaystyle\int_0^\pi f_{(x)}dx$$
Any hints?

Integration by parts tells you that
$$
\int_{0}^{\pi} f(x)\,dx=
\Bigl[xf(x)\Bigr]_0^{\pi}
-\int_0^\pi xf'(x)\,dx
$$
Now compute $f'(x)$ using the fundamental theorem of calculus.
