Evaluating $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$ The question goes:

Evaluate $\int_1^{\pi/2}\frac{\sin(x)\ln(x)}{x^2}dx$.

I have tried by parts, substitutions and other techniques I have learned from first year...but I didn't get any thing look good...
 A: The only viable solution using first-year knowledge is a series solution. We start by replacing the sine function with its respective Taylor series. Then we can integrate term by term
$$I=\int_1^{\frac{\pi}{2}}\frac{\ln{x}\sin{x}}{x^2}dx=\int_1^{\frac{\pi}{2}}\frac{\ln{x}}{x^2}\left(x+\sum_{k=1}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right)dx$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\sum_{k=1}^\infty\frac{(-1)^k}{(2k+1)!}\int_1^{\frac{\pi}{2}}x^{2k-1}\ln{x}dx$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\sum_{k=1}^\infty\frac{(-1)^k}{(2k+1)!}\left(\frac{\ln{\frac{\pi}{2}}}{2}\frac{(\pi^2/4)^k}{k}-\frac{1}{4}\frac{(\pi^2/4)^k}{k^2}+\frac{1}{4}\frac{1}{k^2}\right)$$$$=\frac{1}{2}\ln^2\frac{\pi}{2}+\frac{1}{2}\ln\frac{\pi}{2}\sum_{k=1}^\infty\frac{(-\pi^2/4)^k}{k(2k+1)!}-\frac{1}{4}\sum_{k=1}^\infty\frac{(-\pi^2/4)^k}{k^2(2k+1)!}+\frac{1}{4}\sum_{k=1}^\infty\frac{(-1)^k}{k^2(2k+1)!}$$
Without special functions, this is probably the closest you can get to a solution.
A: In the same spirit as @phi-rate
$$I=\int_1^{\pi/2}\frac{\sin(x)\log(x)}{x^2}dx$$
performing a series expansion around $x=1$
$$\frac{\sin(x)\log(x)}{x^2}=\sum_{n=1}^\infty \frac{a_n}{n!}\,(x-1)^n$$ This gives
$$I=\sum_{n=1}^\infty\frac {a_n}{(n+1)!} \left(\frac{\pi-2}2\right)^{n+1}$$
Using $s=\sin(1)$ and $c=\cos(1)$, the first coefficients are
$$\left(
\begin{array}{cc}
n & a_n \\
 1 & s \\
 2 & 2 c-5 s \\
 3 & 23 s-15 c \\
 4 & 100 c-124 s \\
 5 & 789 s-720 c \\
 6 & 5750 c-5793 s \\
 7 & 48243 s-50911 c \\
 8 & 497096 c-449552 s \\
 9 & 4635705 s-5318136 c \\
 10 & 61946698 c-52414485 s \\
 11 & 644829151 s-781045991 c \\
 12 & 10603366508 c-8575541148 s \\
\end{array}
\right)$$
Using only the above terms leads to $I=\color{red}{0.0731}296$ while the exact value is $I=0.0731869$
Using twice more terms, $I=\color{red}{0.073186}859$ while the exact value is                        $0.073186903$
A: Using the two points Taylor approximation given by @Clerk in comments to this question
$$\sin(x)\sim\left(\frac{5}{\pi ^7}-\frac{1}{2 \pi ^5}\right) x^4 (x-\pi )^4+\left(\frac{1}{6 \pi ^3}-\frac{2}{\pi ^5}\right) x^3 (x-\pi )^3+\frac{x^2 (x-\pi )^2}{\pi ^3}-\frac{x (x-\pi )}{\pi }$$  whose error is $7.5\times 10^{-5}\,z^5$ and the fact that
$$\int_1^{\frac \pi 2}x^n \, \log(x)_,dx=\frac 1{(n+1)^2 }\left(1+\left(\frac{\pi}{2 }\right)^{n+1} \left((n+1) \log
   \left(\frac{\pi }{2}\right)-1\right)\right)$$ we obtain
$$I=\int_1^{\pi/2}\frac{\sin(x)\log(x)}{x^2}dx=\color{red}{0.07318}539$$ while the exact value is                                             $0.07318690$
