<s>[INST] 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...
 [/INST] 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.
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