Problem source

Evaluate  $\int_0^{{\pi}} x \sin x\,\d x$ and $\int_0^{{\pi}} x \cos x\,\d x\;$.
The function $\f$ satisfies the equation
\begin{equation}
\f(t)=t + \int_0^{{\pi}} \f(x)\sin(x+t)\,\d x\;. 
\tag{$*$}
\end{equation}
Show that 
\[
\f(t)=t + A\sin t + B\cos t\;,        
\]
where $A= \int_0^{{\pi}}\,\f(x)\cos x\,\d x\;$ and $B= \int_0^{{\pi}}\,\f(x)\sin x\,\d x\;$. 
Find $A$ and $B$ by substituting for $\f(t)$ and $\f(x)$ in $(*)$
and equating coefficients of $\sin t$ and $\cos t\,$.

Solution source

\begin{align*}
&& I &= \int_0^\pi x \sin x \d x \\
&&&= \left [ -x \cos x  \right]_0^\pi + \int_0^{\pi} \cos x \d x \\
&&&= \pi \\
\\
&& J &= \int_0^\pi x \cos x \d x \\
&&&= \left [ x \sin x \right]_0^\pi - \int_0^\pi \sin x \d x \\
&&&= -2
\end{align*}

\begin{align*}
&& f(t) &= t + \int_0^\pi f(x) \sin (x+t) \d x \\
&&&= t + \int_0^\pi f(x) \left ( \sin t \cos x  + \cos t \sin x \right) \d x \\
&&&= t + \sin t \int_0^{\pi} f(x) \cos x \d x + \cos t \int_0^{\pi} f(x) \sin x \d x \\
\\
&& A &= \int_0^\pi (x + A \sin x + B \cos x) \cos x \d x \\
&&&= -2+ \frac{\pi}{2} B \\
&& B &= \int_0^{\pi} (x + A \sin x + B \cos x ) \sin x \d x \\
&&&= \pi + \frac{\pi}{2} A \\
\Rightarrow && (A,B) &= (-2,0)
\end{align*}