Let $B = \int_{0}^{\pi/4} cos(x)\cdot ln(x)$
rewrite the integral $B$ so that
$B = a - U $
'where $a \in \mathbb{R}$ is a constant and $U = \int_{0}^{\pi/4}g(x)dx$. The function $g(x)$ has no singularity at $0$ ,that is, it is continuous and bounded around zero.'
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My solution to the problem is to use partial integration so that
$[sin(x)ln(x)]_{t}^{\pi/4} - \int_{0}^{\pi/4}\dfrac{sin(x)}{x}dx$
Since $lim_{t->0^+}sin(x)\cdot ln(x)=0$ we therefore get $a = sin(\pi/4)\cdot ln(\pi/4)$.
$g(x) = \dfrac{sin(x)}{x}$ is 'continuous and bounded around zero' however it is not defined at zero. Is it singular at zero? is the textbook being unclear of what they want? I believe that $\dfrac{sin(x)}{x}$ is singular at zero since it is not defined there.
The function $g(x) = \dfrac{sin(x)}{x}$ can be continuously extended to the same function with $g(0)=1$, which is possible because $\lim_{x\to 0}\dfrac{sin(x)}{x}=1$. Now it's continuous and bounded around $0$.
Extra babble:
$U$ here could also be interpreted as an improper Riemann integral without extending the $g$ you found: $U=\lim_{t\downarrow 0}\int_{t}^{\pi/4}g(x)dx$.
For this improper Riemann integral, $g(x)$ doesn't have to be defined in $x=0$, but that's not what the question is asking for.